Hilbert's Nullstellensatz without Axiom of Choice 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. 
 A: 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.
