Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951.
Can we prove the following theorem without Axiom of Choice?
Theorem Let $A$ be a commutative algebra of finite type over a field $k$. Let $I$ be an ideal of $A$. Let $\Omega(A)$ be the set of maximal ideals of $A$. Let $V(I)$ = {$\mathfrak{m} \in \Omega(A)$; $I \subset \mathfrak{m}$}. Let $f$ be any element of $\cap_{\mathfrak{m} \in V(I)} \mathfrak{m}$. Then there exists an integer $n \geq 1$ such that $f^n \in I$.
EDIT So what's the reason for the downvotes?
EDIT What's wrong with trying to prove it without using AC? When you are looking for a computer algorithm for solving a mathematical problem, such a proof may provide a hint. At least, you can be sure that there is a constructive proof.
EDIT To Martin Brandenburg, I think this thread also answers your question.
Lemma 1 Let $A$ be a commutative algebra of finite type over a field $k$. Then there exists a maximal ideal $P$ of $A$ such that $A/P$ is a finite $k$-module.
Proof: If every element of $A$ is algebraic over $k$, then $A$ is a finite $k$-module. Therefore the lemma is trivial. Hence we can assume otherwise. By Noether normalization lemma (this can be proved without AC), there exist algebraically independent elements $x_1,\dots, x_n$ in $A$ such that $A$ is a finitely generated module over the polynomial ring $A' = k[x_1,\dots, x_n]$. Let $\mathfrak{m} = (x_1,\dots, x_n)$ be the ideal of $A'$ generated by $x_1,\dots, x_n$. Clearly $\mathfrak{m}$ is a maximal ideal of $A'$. By the answer by QiL to this question, there exists a prime ideal $P$ of $A$ lying over $\mathfrak{m}$. Since $A/P$ is a finitely generated module over $k = A'/m$, $P$ is a maximal ideal. QED
Lemma 2 Let $A$ be a commutative algebra of finite type over a field k. Let $f$ be a non-nilpotent element of $A$. Then there exists a maximal ideal $P$ of $A$ such that $f \in A - P$.
Proof: Let $S$ = {$f^n; n = 1, 2, \dots$}. Let $A_f$ be the localization with respect to $S$. By Lemma 1, there exists a maximal ideal $\mathfrak{m}$ of $A_f$ such that $A_f/\mathfrak{m}$ is a finite $k$-module. Let $P$ be the inverse image of $\mathfrak{m}$ by the canonical morphism $A \rightarrow A_f$. $A/P$ can be identified with a subalgebra of $A_f/\mathfrak{m}$. Since $A_f/\mathfrak{m}$ is a finite $k$-module, $A/P$ is also a finite $k$-module. Hence $P$ is a maximal ideal. Clearly $f \in A - P$. QED
The title theorem follows immediately from the following lemma by replacing $A$ with $A/I$.
Lemma 3 Let $A$ be a commutative algebra of finite type over a field k. Let $\Omega(A)$ be the set of maximal ideals of $A$. Let $f$ be any element of $\cap_{\mathfrak{m} \in \Omega(A)} \mathfrak{m}$. Then $f$ is nilpotent.
Proof: This follows immediately from Lemma 2.