positive depth ring has regular minimal generator

The associated primes are finite, none is $\mathfrak{m}$, and their union is the set of zero divisors. Discard primes contained in others to obtain an antichain ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_s$. Prime avoidance chooses $a\in \mathfrak{m}$ outside their union (if the list is empty this restriction is vacuous). If $a\mathrm{\notin }{\mathfrak{m}}^2$, take $x=a$.

If $a\in {\mathfrak{m}}^2$, Nakayama and positive depth give $\mathfrak{m}\mathrm{\ne }{\mathfrak{m}}^2$; choose $b\in \mathfrak{m}\setminus {\mathfrak{m}}^2$. If $b$ avoids every retained prime take $x=b$. Otherwise divide them into the nonempty class $T$ containing $b$ and the class $U$ not containing it. For each $\mathfrak{p}\in U$, antichain incomparability and prime avoidance give $c_{\mathfrak{p}}\in \mathfrak{p}$ outside all primes of $T$. Put $c={\prod }_{\mathfrak{p}\in U}{c}_{\mathfrak{p}}$, with empty product $1$.

Then $x=b+ac$ is outside ${\mathfrak{m}}^2$, since $ac\in {\mathfrak{m}}^2$. At a prime of $T$, $b$ lies in the prime and $ac$ does not. At a prime of $U$, $ac$ lies in the prime and $b$ does not. Thus $x$ avoids every associated prime and is a nonzerodivisor. No infinite-field argument was used.