 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 $R_f\ne 0$, 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,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