The equation must define the intended scheme

Let $R=k[x]$, so every element of $R$ is a finitely supported coefficient sequence ${\sum }_j{c}_j{x}^j$ with unique coefficients $c_j\in k$ by [F1]; let $I=(x^p)$ and $J=(x)$ be the principal ideals generated by $x^p$ and $x$ [F5], so that $I=\{x^{p}h:h\in R\}$, $J=\{xh:h\in R\}$ and $I\subseteq J$ because $x^p=x\cdot x^{p-1}$; let $A=R\mathrm{/}I$ and $B=R\mathrm{/}J$ be the quotient rings with classes $\tau =x+I$ and $\sigma =x+J$.

The evaluation map $\epsilon :R\to k$ with $\epsilon (x)=0$ exists and is unique by [F2]; its kernel is the maximal ideal $(x)$ by [F15], so $\epsilon$ kills the ideals $I\subseteq J=(x)$ and, by [F4], induces surjective $k$-algebra homomorphisms $\bar{\epsilon }:A\to k$ with $\tau \mapsto 0$ and ${\epsilon }_1:B\to k$ with $\sigma \mapsto 0$; by [F16] these are $k$-rational points $0\in X_p(k)$ and $0\in X_1(k)$ of $X_p=\mathrm{Spec}A$ and $X_1=\mathrm{Spec}B$, and ${\tau }^p=0$ in $A$.

Division by the monic polynomial $x^p$ in $R$ [F3] gives every $f\in R$ a unique $f=q{x}^p+r$ with $r=0$ or $\mathrm{deg}r<p$; comparing coefficients in the unique coefficient representation [F1] shows that the classes $1,\tau ,\dots ,{\tau }^{p-1}$ form a $k$-basis of $A$, that $1\mathrm{\ne }0$ and $\tau \mathrm{\ne }0$ in $A$, and that $A=k\cdot 1\oplus (\tau )$ as $k$-vector spaces, since $(\tau )$ is spanned by $\tau ,\dots ,{\tau }^{p-1}$ and every $a={\sum }_{j<p}{c}_j{\tau }^j$ has the unique decomposition $a=c_0+\tau \cdot {\sum }_{0<j<p}{c}_j{\tau }^{j-1}$; here $p\ge 2$ by [F21].

Division by the monic polynomial $x$ [F3] gives every $f\in R$ the unique expression $f=qx+f(0)$; hence every class in $B=R\mathrm{/}(x)$ is the class of a unique constant, the structure map $k\to B$ is an isomorphism, $\sigma =0$ in the field $B$, and the unique prime ideal of $B$ is $(0)$ by [F6].

Tangent space of the thickened scheme: the Jacobian matrix of the single equation $x^p$ at the point $0$ is the $1\times 1$ matrix $(p{x}^{p-1}{\mathrm{\mid }}_{x=0})$ by [F18]; in characteristic $p$ the coefficient $p\cdot 1_k$ is $0$ by [F20], so $p{x}^{p-1}$ is the zero polynomial and the matrix is $(0)$; by [F19] there is a canonical $k$-linear isomorphism $T_0{X}_p\cong \ker (0:k\to k)=k$, of dimension $1$.

Finite type over $k$: the rings $A=k[x]\mathrm{/}(x^p)$ and $B=k[x]\mathrm{/}(x)$ are quotients of the polynomial ring $k[x]$, hence of finite type over $k$ by [F23]; over the affine base $\mathrm{Spec}k$ each of the schemes $X_p$ and $X_1$ is covered by its single affine chart with a finite-type coordinate ring, so [F24] makes the structure morphisms $X_p\to \mathrm{Spec}k$ and $X_1\to \mathrm{Spec}k$ of finite type.

The constant-term map: by step 2.1 the assignment $A\to k$, ${\sum }_{j<p}{c}_j{\tau }^j\mapsto c_0$, is the composite of the quotient map $A\to A\mathrm{/}(\tau )$ with the inverse of the isomorphism $k\to A\mathrm{/}(\tau )$ induced by the constants, which is injective because $k\cdot 1\cap (\tau )=0$ in the direct sum $A=k\cdot 1\oplus (\tau )$; it is a surjective ring homomorphism with kernel $(\tau )$, so $A\mathrm{/}(\tau )\cong k$ is a field and $(\tau )$ is a maximal ideal by [F7] and a prime ideal by [F8].

$A$ is Noetherian: it is a commutative $k$-algebra whose underlying $k$-vector space has the finite basis $1,\tau ,\dots ,{\tau }^{p-1}$ of step 2.1, so [F14] makes $A$ a Noetherian ring, and $X_p=\mathrm{Spec}A$ is locally Noetherian by [F13].

The two ideals have the same reduced zero set: $(x^p)\subseteq (x)$ by step 1.1; the ideal $(x)$ is maximal, hence prime, by [F15] and [F8]; if $f^n\in (x^p)\subseteq (x)$ for some $n\ge 1$ then primeness of $(x)$ [F6] gives $f\in (x)$, so $\sqrt{(x^p)}\subseteq (x)$; conversely $x^p\in (x^p)$ gives $x\in \sqrt{(x^p)}$ and hence $(x)\subseteq \sqrt{(x^p)}$; therefore $\sqrt{(x^p)}=(x)=\sqrt{(x)}$ by [F22], while the inclusion is strict, $x\mathrm{\notin }(x^p)$, since otherwise $\tau =x+(x^p)=0$ against step 2.1.

Every prime ideal $\mathfrak{p}$ of $A$ contains $\tau$: since ${\tau }^p=0\in \mathfrak{p}$ by step 1.2, induction on the exponent using primeness [F6] gives $\tau \in \mathfrak{p}$; hence $(\tau )\subseteq \mathfrak{p}$, and maximality of $(\tau )$ from step 3.1 forces $\mathfrak{p}=(\tau )$; therefore $(\tau )$ is the unique prime ideal and, being maximal, the unique maximal ideal, so $A$ is a nonzero local ring with residue field $A\mathrm{/}(\tau )\cong k$ and $X_p=\mathrm{Spec}A$ has exactly one point.

The Krull dimension is $\dim A=0$: $A$ is nonzero by step 2.1 and its only prime ideal is $(\tau )$ by step 4.1, so the only strict chains of prime ideals of $A$ have length $0$; by [F10], $\dim A=0$.

Embedding dimension: for the maximal ideal $\mathfrak{m}=(\tau )$ of the local ring $A$ one has ${\mathfrak{m}}^2=({\tau }^2)$ and $\mathfrak{m}=k\cdot \tau +{\mathfrak{m}}^2$, because for $a=c_0+\tau b$ from the decomposition of step 2.1 the product $a\tau =c_0\tau +{\tau }^{2}b$ lies in $k\cdot \tau +{\mathfrak{m}}^2$; moreover $\tau \mathrm{\notin }{\mathfrak{m}}^2$, since every element of $({\tau }^2)$ has zero coordinate of $\tau$ in the basis of step 2.1 while $\tau$ has coordinate $1$; hence the class of $\tau$ is a $k$-basis of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ and $\mathrm{edim}A={\dim }_k(\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2)=1$ by [F11].

Local rings of the two points: by [F17] the stalk of $X_p$ at its unique point, the prime $\mathfrak{m}=(\tau )$, is ${\mathcal{O}}_{X_p,0}\cong A_{\mathfrak{m}}$; every $s\in A\setminus \mathfrak{m}$ has nonzero constant term $c_0$ in the decomposition of step 2.1, so $s=c_0(1+u)$ with $u=c_0^{-1}\tau b$ and $u^p=c_0^{-p}{\tau }^p{b}^p=0$, and $1+u$ is a unit with the explicit inverse ${\sum }_{j=0}^{p-1}(-u{)}^j$; hence every $s\mathrm{\notin }\mathfrak{m}$ is a unit of $A$ and the localisation map $A\to A_{\mathfrak{m}}$ is an isomorphism (surjective because $a\mathrm{/}s=a{s}^{-1}$, injective because $ta=0$ for a unit $t\mathrm{\notin }\mathfrak{m}$ forces $a=0$), so ${\mathcal{O}}_{X_p,0}\cong A$ by [F9]; for the field $B$ the unique prime is $(0)$ by step 2.2, its localisation is $B$ itself and ${\mathcal{O}}_{X_1,0}\cong B\cong k$.

The local ring $A$ is not regular: it is nonzero, commutative and Noetherian by steps 2.1, 3.2, and $\mathrm{edim}A=1\mathrm{\ne }0=\dim A$ by steps 5.1 and 5.2, so [F11] gives that $A$ is not a regular local ring.

Tangent space and regularity of the reduced point: the Jacobian of the equation $x$ is the matrix $(1)$, so [F19] gives $T_0{X}_1\cong \ker (1:k\to k)=0$; the ring $B\cong k$ is a field, hence a nonzero Noetherian local ring whose maximal ideal is $0$ and whose only prime is $(0)$, so $\dim B=0$ by [F10] and $\mathrm{edim}B={\dim }_{k}0=0$ by [F11], making $B$ regular local; therefore ${\dim }_k{T}_0{X}_1=0=\dim B=\dim {\mathcal{O}}_{X_1,0}$ and $0$ is a regular point of $X_1$ by [F12], while ${\dim }_k{T}_0{X}_p=1\mathrm{\ne }0=\dim A=\dim {\mathcal{O}}_{X_p,0}$ by steps 5.1, 6.1 and 5.3, so $0$ is not a regular point of $X_p$ by [F12] and [F13].

Smoothness form under AC: assume [F28]; by [F25] a finite-type $k$-scheme is smooth over $k$ exactly when every local ring of every base change along a field extension $K\mathrm{/}k$ is regular. For $X_1\cong \mathrm{Spec}B$ and any field extension $K\mathrm{/}k$, the base change is covered by the single affine chart $\mathrm{Spec}(B{\otimes }_{k}K)$ by [F26], and $B{\otimes }_{k}K\cong K[x]\mathrm{/}(x)\cong K$ by [F27], a field, whose maximal ideal $0$ gives a regular local ring; hence $X_1$ is smooth over $k$. For $X_p$ the field extension $K=k$ gives the base change $X_p$ itself by [F26], whose local ring at $0$ is $A$, not regular by step 6.1, so $X_p$ is not smooth over $k$. The comparisons exhibit the single extension $K=k$ and the single points involved, so AC is used only through the criterion [F25] of [F28] and no dependent choice is made.

Conclusion: the ideals $(x^p)$ and $(x)$ have the same reduced zero set, the single point $0$ of ${\mathbb{A}}_k^1$, yet at that point the tangent space $T_0{X}_p\cong k$ is one-dimensional while $T_0{X}_1=0$ is zero-dimensional by steps 2.3 and 7.1, and $0$ is nonregular on the thickened scheme $X_p$ while regular on the reduced point $X_1$ by step 7.1; so the same reduced zero set does not determine the tangent space or the singular locus, and the false claim stated above is refuted; the vanishing Jacobian of $x^p$ in characteristic $p$ is a correct statement about the nonreduced scheme $X_p=\mathrm{Spec}(k[x]\mathrm{/}(x^p))$ that the equation $x^p=0$ defines, and since no reduction is performed the nilpotent $\tau \mathrm{\ne }0$ with ${\tau }^p=0$ is retained throughout.

Boundary and scope dispositions: both schemes are nonempty, each having exactly the one point exhibited in steps 1.2, 2.2 and 4.1; the Krull dimension is $0$ in both cases (steps 5.1 and 7.1) while the tangent dimensions are $1$ and $0$, and the zero polynomial $p{x}^{p-1}$ of step 2.3 is the identically zero Jacobian; each system has one variable, one equation and a single principal generator (step 1.1), and the least prime $p=2$ is included, where ${\mathfrak{m}}^2=({\tau }^2)=0$ and the class of $\tau$ still spans $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ because $\tau \mathrm{\ne }0$ by step 2.1; the example is the degenerate nonreduced case ${\tau }^p=0$ with $\tau \mathrm{\ne }0$, and no reduction of $I$ to its radical is performed, which is exactly why ${\dim }_k{T}_0{X}_p=1$ while the reduced point has zero tangent space; the field $k$ is an arbitrary field of characteristic $p>0$, with no algebraic closure, perfectness or finiteness hypothesis, and the phenomenon is characteristic $p$ because only then is the derivative $p{x}^{p-1}$ of $x^p$ the zero polynomial; the refutation of the false claim is an instance of failure of the implication "same reduced zero set gives the same tangent space", the smoothness criterion of step 7.2 is used in both directions as stated there, and the Axiom of Choice is confined to that step.