This has a very easy proof if one generalizes it to (infinite-dimensional) linear algebra and forgets about commutative algebra. Let $K/F$ be an extension of fields (e.g., could take $F$ to be a prime field), $V$ an $F$-vector space (such as a polynomial ring over a prime field), and $W$ a $K$-subspace of $V_K := K \otimes_F V$.

Among all subfields $K_0$ of $K$ over $F$ such that $W = K \otimes_{K_0} W_0$ for a (visibly unique) $K_0$-subspace $W_0$ of $V_{K_0}$, we claim that the intersection of these fields works too.

(In case $V$ is an $F$-algebra and $W$ is an ideal of $V_K$, obviously $W_0$ is an ideal of $V_{K_0}$, so this really does imply Weil's result. In fact, it gives a more general result: no need to assume the algebras are finitely generated.)

Proof: Choose an $F$-basis $\{v_i\}_{i \in I}$ of $V$, so there is a subset $J$ of $I$ such that $\{v_j \bmod W\}_{j \in J}$ is a $K$-basis of $V/W$. For $i' \in I - J$, expand $v_{i'} \bmod W \in V/W$ in this basis:

$$v_{i'} \equiv \sum_{j \in J} a_{i'j} v_j \bmod W$$
with $a_{i'j} \in K$. The necessary and sufficient condition on $K_0$ for $W_0$ to exist is that $K_0$ contains every $a_{i'j}$ (for $j \in J$ and $i' \in I - J$). So the subfield $F(a_{i'j})_{i', j}$ is the desired minimal subextension of $K$ over $F$. QED

There is a very elegant modern discussion of the theme of field of definition (for closed subschemes, morphisms, etc.) without any finiteness hypotheses in EGA IV$_2$, 4.8.