= Solution
The weak <Hilbert Nullstellensatz> says that if a field is a finitely generated algebra over $k$, then it is a finite algebraic extension of $k$. Since every maximal ideal is prime, the nilradical $\mathcal N(R)$ is contained in the <Jacobson radical> $J(R)$.
Conversely, let $f$ be nonnilpotent. Then $R_f\ne0$, and it is a finitely generated $k$-algebra, so it has a maximal ideal $\mathfrak n$. Its contraction $\mathfrak p$ to $R$ avoids $f$. Moreover,
$$
R/\mathfrak p\hookrightarrow R_f/\mathfrak n
$$
is a finitely generated $k$-domain inside a finite algebraic extension of $k$, hence is itself a field. Thus $\mathfrak p$ is maximal and does not contain $f$. Therefore $f\notin J(R)$, proving
$$
\boxed{\mathcal N(R)=J(R).}
$$
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