= Solution
The statement is true. Choose a <maximal ideal> $\mathfrak m\subset B$ and write $L=B/\mathfrak m$, a <field extension> of $k$. Taking the quotient by $A\otimes_k\mathfrak m$ gives
$$
(A\otimes_kB)/(A\otimes_k\mathfrak m)
\cong A\otimes_kL,
$$
so $A\otimes_kL$ is a <finitely generated algebra> over $k$, and therefore over $L$.
By <finite generation descends along a field extension>, $A$ is finitely generated over $k$. Indeed, collect the finitely many coefficients from $A$ occurring in a finite set of $L$-algebra generators of $A\otimes_kL$. If $A_0$ is the $k$-subalgebra generated by those coefficients, then $(A/A_0)\otimes_kL=0$; because a field extension is a <faithfully flat module>, $A=A_0$. Interchanging $A$ and $B$ proves that $B$ is also finitely generated.
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