A proof of the proposed result that is similar (if not identical) to [Peter Kropholler's](https://mathoverflow.net/a/430392/84349) and YCor's proofs can be derived from two *well-known* results, namely **Lemma 2** and **Theorem 3** below, together with the **Artin-Rees lemma** (used in Claim 5).
We shall establish:
> **Claim 1.** Let $R$ be a commutative unital ring. Let $\mathcal{I}$ be intersection of all ideals $I$ of $R$ such that $R/I$ is a local ring of finite cardinal. 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](https://en.wikipedia.org/wiki/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 [nilradical](https://en.wikipedia.org/wiki/Nilradical_of_a_ring) of $R$. Consequently, there is $n \ge 0$ 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.
The following result mentioned by YCor is instrumental.
(It relies on the Artin-Rees Lemma as indicated by Peter Kropholler).
> **Claim 5.** Let $R$ be a commutative unital Noetherian ring. Then $R$ is residually local, i.e.,
for every non-zero $x \in R$ there is an ideal $I$ of $R$ such that $x \notin I$ and $R/I$ is local. (In other words, the intersection of all ideals $I$ such that $R/I$ is local, results in the null ideal.)
> *Proof.* 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 $\overline{x} = x + I$. By construction, we know that $\overline{x}$ is contained in every non-zero ideal of $\overline{R}$. It also follows from our assumptions on $x$ and $I$ that $\overline{R}\overline{x}$ is a simple $\overline{R}$-module, so that the annihilator $M$ of $\overline{x}$ is a maximal ideal of $\overline{R}$. We claim that there is $n \ge 1$ such that $M^n = \{0\}$. If the claim holds true, then any maximal ideal of $\overline{R}$ contains a power of $M$ and hence is equal to $M$, which shows that $\overline{R}$ is local.
Reasoning by way of contradiction, we assume that $M^n \neq \{0\}$ for every $n \ge 1$. As $\overline{R}$ is Noetherian, we can apply the Artin-Rees lemma [Theorem 8.5, 3]. This lemma yields a positive integer $c$ such that $M^n \cap \overline{R} \overline{x} = M^{n - c}(M^c \cap \overline{R} \overline{x})$ for every $n > c$. Taking $n = c + 1$, we obtain that $\overline{R} \overline{x} = M \overline{R} \overline{x} = \{0\}$, a contradiction.
Now we are in position to prove Claim 1.
> *Proof of Claim 1.* Combine Claims 4 and 5.
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[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.