A nilpotent thickening of an affine scheme is affine

First suppose $I^2=0$. The sheaf $I$ is a quasi-coherent module on $Y$, because $I$ annihilates itself. The closed immersion does not change the underlying topological space. Thus [F1] gives $H^1(X,I)=H^1(Y,I)=0$, and the exact sequence $0\to I\to {\mathcal{O}}_X\to {\mathcal{O}}_Y\to 0$ gives a surjection $A=\mathrm{\Gamma }(X,{\mathcal{O}}_X)\to B=\mathrm{\Gamma }(Y,{\mathcal{O}}_Y)$ with square-zero kernel.

Around each point choose an affine open $V\subset X$, and then a principal open $D(b)\subset Y=\mathrm{Spec}B$ contained in $V\cap Y$. Lift $b$ to $a\in A$ by step 1.1. Its nonvanishing open $X_a$ has underlying space $D(b)$, lies in $V$, and is the principal open of the restriction of $a$ to $V$, hence affine. By [F2], $\mathrm{\Gamma }(X_a,\mathcal{O})=A_a$: surjectivity follows by clearing powers of $a$, and a section of $A$ zero on $X_a$ is annihilated by a power of $a$, by the same extension/localization argument on the finite affine cover of $X$. Choose finitely many of these opens covering $X$. Their corresponding $D(a)\subset \mathrm{Spec}A$ cover that spectrum, since $\mathrm{Spec}A$ and $\mathrm{Spec}B$ have the same underlying space under the square-zero quotient. The canonical map $X\to \mathrm{Spec}A$ therefore is an isomorphism on this affine-open cover, and hence globally.

For general $I^m=0$, start with $X_1=Y$, and successively thicken to the closed schemes $X_j$ defined by $I^j$, for $j=2,\dots ,m$. The ideal of $X_{j-1}$ in $X_j$ is $I^{j-1}\mathrm{/}{I}^j$, whose square is zero because $2(j-1)\ge j$. Steps 1.1 and 2.1 show inductively that each $X_j$ is affine; $X_m=X$. AC is inherited from the cohomology and section-extension suppliers.