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.
