Let $R$ be a finitely generated $\mathbb{Z}$-algebra. I want to show that its Jacobson radical $J(R) = 0.$ I know that if $Q$ is a maximal ideal of $R$ then $R/Q$ is finite. The question first asks to show that if $0 \neq b \in R$, and $Q$ is a maximal ideal of $S = R[b^{-1}]$, then $P = Q \cap R$ is also maximal.
I know that $P$, being the contraction of a prime ideal, must be a prime ideal that doesn't meet the set $Y = \{b^n : n \in \mathbb{N}\}$ (by properties of localization). Let $r \in R \setminus P.$ Then $P+(r)$ is an ideal strictly containing $P$, so the extension of it into $S$, which is the ideal $e(P+(r)) = e(P)+e((r)),$ must contain $e(P) = Q$ as extensions preserve inclusions and sums. But if $e(P)+e((r)) = Q$, we have $e((r)) \subseteq e(P),$ hence $ce((r)) \subseteq ce(P)=P,$ where $c$ is the contraction map. But for any ideal, $ce(I) \supseteq I$, so we have $P \supseteq(r),$ a contradiction.
So $e(P+(r)) = S$, which means $P+(r)$ meets $Y = \{b^n : n \in \mathbb{N}\}$. If $P$ were not maximal, there would be some $r \in R$ such that $P+(r)$ is a proper ideal, and then there would be a prime ideal containing $P+(r),$ which is impossible because $P+(r)$ meets $Y$. Hence $P$ is maximal.
Now I want to show that $J(R)$ is trivial. Suppose there is a nonzero $b \in J(R).$ Then we can define $R[b^{-1}]$ as above. A maximal ideal here then descends to a maximal ideal of $R$ not containing $b$, a contradiction. So $J(R) = 0$.
I believe this solution is correct, but I never used the fact that $R$ is a finitely generated $\mathbb{Z}$-algebra, which makes me think I did something wrong. Is there a flaw in my argument or is there perhaps a simpler way to make this work?
This is not a contradiction. Yes, $P+(r)$ meets $Y$, but so what? There is nothing wrong with having a prime ideal in $R$ that meets $Y$. Indeed, if $b$ is not a unit in $R$, then there must exist prime ideals in $R$ that meet $Y$: just take any maximal ideal containing $b$.
To show $P$ is maximal, you can instead consider the induced inclusion $R/P\to S/Q$. Since $S/Q$ is a finite field (by the fact you mentioned in your first paragraph) and every subring of a finite field is a field, $R/P$ is also a field, so $P$ is maximal.