Let $R$ be a commutative, finitely generated $\mathbb{Z}$-algebra, then the nil radical is equal to the Jacobson radical.

I am not able to make much traction on this, nor can I find this result in any book I've look at. So far I've reasoned that since $R \cong \mathbb{Z}[x_1,\dots,x_n]/I$ for some ideal $I$, then $nil(R) = \sqrt{I}/I$ and $J(R) = M/I$, where $M$ is the intersection of all ideals containing $I$. Clearly, the jacobson radical contains all nilpotent elements, due the to the quasiregular condition, but I can't seem to follow through on the converse.

I already know that if $R$ is a field, then it must be finite, but I can't see how this would help me prove the general case.

Any help would be greatly appreciated.