Update: there is an error in my proof of Claim 1 below. (Looking for a fix.)
A proof of the proposed result that is similar (if not identical) to Peter Kropholler's can be derived from two well-known results, namely Lemma 2 and Theorem 3 below.
We shall establish a statement which is actually equivalent to OP's result:
Claim 1. Let $R$ be a commutative unital ring. Let $\mathcal{I}$ be intersection of all ideals $I$ of $R$ such $R/I$ is local. If $R$ is a finitely generated $\mathbb{Z}$-algebra, then $\mathcal{I}$ is $\{0\}$.
The main results we need are:
Lemma 2. [Lemma 4.8, 1]. A field which is finitely generated as a ring is finite.
Theorem 3. [Theorem 4.19 (Nullstellensatz, General form), 2]. Let $R$ be a Jacobson ring and let $S$ be a finitely generated $R$-algebra. Then $S$ is a Jacobson ring.
As an intermediate step, we shall prove:
Claim 4. Let $R$ be a finitely generated $\mathbb{Z}$-algebra. If $R$ is local, then $R$ is a finite ring.
Proof. Since $R$ is Noetherian, its unique maximal ideal $\mathfrak{m}$ is finitely generated. As $R$ is Jacobson by Theorem 2, the ideal $\mathfrak{m}$ is also the nil-radical of $R$. There is therefore $n \ge 0$ such that $\mathfrak{m}^n = 0$, which shows in particular that $R$ is Artinian. To conclude, it only remains to show that the residual field $R/\mathfrak{m}$ of $R$ is finite, which is given by Lemma 1.
Now we are in position to prove Claim 1.
Proof of Claim 1. Let $x \in R \setminus \{0\}$ and let $I$ be an ideal of $R$ maximal among the ideals of $R$ not containing $x$. Such an $I$ exists by Zorn's lemma. We shall prove that $\overline{R} = R/I$ is local. We can assume, without loss of generality, that $\overline{R}$ is not a field. Let $\overline{x} = x + I$. We claim that $\overline{x}^2 = 0$. By hypothesis, $\overline{x} \neq 0$ is contained in any non-trivial ideal of $\overline{R}$, but also in all maximal ideals of $\overline{R}$. Hence $\overline{x}$ belongs to the Jacobson radical of $\overline{R}$ so that $\overline{x}^2 = 0$ by Nakayama's lemma[Theorem 2.2, 3]. We claim that $\overline{x} M = \{0\}$ for every maximal ideal $M$ of $\overline{R}$. Reasonning by way of contradiction, we assume that $\overline{x} M \neq \{0\}$. We have thus $\overline{x} M = M$. As $\overline{R}$ is Noetherian, the ideal $M$ is finitely generated so that the multiplication by $\overline{x}$ induces an isomorphism of $M$ [Theorem 2.4, 3]. But this cannot be true since $\overline{x}^2 = 0$ and $\overline{x} \in M$. Assume now that there are two distinct maximal ideals $M, M'$ of $\overline{R}$. Then $ \overline{x} \overline{R} = \overline{x} (M + M') = \overline{0}$, a contradiction. As a result, the ring $\overline{R}$ is local.
[1] R. Swan, "Excision in algebraic K-theory", 1971.
[2] D. Eisenbud, "Commutative Algebra with a View Towards Algebraic Geometry", 1995.
[3] H. Matsumura, "Commutative Ring Theory", 1989.