Update: My proof of Claim 1 below is incomplete. (Still trying to fill the gap.)
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. Let $\overline{x} = x + I$. Since $\overline{R}\overline{x}$ is a simple $\overline{R}$-module by construction, the annihilator $M$ of $\overline{x}$ is a maximal ideal of $\overline{R}$. Let $\mathcal{Z}$ be the set of zero divisors of $\overline{R}$. We claim that $M = \mathcal{Z}$. The inclusion $M \subseteq \mathcal{Z}$ is obvious. Let $z \in \mathcal{Z}$ and let $y \in \overline{R} \setminus \{0\}$ such that $zy = 0$. As $\overline{x} \in \overline{R}y$, we have $z\overline{x} = 0$, which shows that the reverse inclusion holds. To be continued.
[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.