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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 of $R$. Consequently, there is $n \ge 1$ 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 $\overline{R} \overline{x} = M \overline{R} \overline{x} = \{0\}$, which is the desired contradiction. Observe indeed that $M^c$ and $M^{c + 1}$ are non-zero by assumption, so that both ideals contain $\overline{x}$.

By construction, the ideal $\overline{R}\overline{x} \simeq \overline{R}/M$ is an essential $\overline{R}$-submodule of $\overline{R}$. Therefore the injective hull $E(\overline{R}\overline{x}) \simeq E(\overline{R} / M)$ contains $\overline{R}$. In particular, $1 \in E(\overline{R}\overline{x})$, so that $M^n \cdot 1 = \{0\}$ for some $n \ge 1$ by [Theorem 18.4.v, 3]. Thus $M^n = \{0\}$, which implies that $\overline{R}$ is local, as desired.

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 of $R$. Consequently, there is $n \ge 1$ such that $\mathfrak{m}^n = 0$, which shows in particular that $R$ is Artinian ($R$ is zero-dimensional and Noetherian). 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 can be proved by means of the Artin-Rees lemma as indicated by Peter Kropholler, or alternatively by using a result of Matlis'theory of injective modules of Noetherian rings [Theorem 18.4.v, 3].

Claim 5. Let $R$ be a commutative unital Noetherian ring. Then $R$ is residually local and Artinian, 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 and Artinian. (In other words, the intersection of all ideals $I$ such that $R/I$ is local and Artinian, 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 prime ideal of $\overline{R}$ contains a power of $M$ and hence is equal to $M$, which shows that $\overline{R}$ is local and Artinian. 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 $\overline{R} \overline{x} = M \overline{R} \overline{x} = \{0\}$, which is the desired contradiction. Observe indeed that $M^c$ and $M^{c + 1}$ are non-zero by assumption, so that both ideals contain $\overline{x}$.

By construction, the ideal $\overline{R}\overline{x} \simeq \overline{R}/M$ is an essential $\overline{R}$-submodule of $\overline{R}$. Therefore the injective hull $E(\overline{R}\overline{x}) \simeq E(\overline{R} / M)$ contains $\overline{R}$. In particular, $1 \in E(\overline{R}\overline{x})$, so that $M^n \cdot 1 = \{0\}$ for some $n \ge 1$ by [Theorem 18.4.v, 3]. Thus $M^n = \{0\}$, which implies that $\overline{R}$ is local and Artinian, as desired.

A proof of the proposed result that is similar (if not identical) to Peter Kropholler's 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).

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 $\overline{R} \overline{x} = M \overline{R} \overline{x} = \{0\}$, which is the desired contradiction. Observe indeed that $M^c$ and $M^{c + 1}$ are non-zero by assumption, so that both ideals contain $\overline{x}$.

A proof of the proposed result that is similar (if not identical) to Peter Kropholler's 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 (alternatively [Theorem 18.4.v, 3]; see proof of Claim 5 and subsequent note).

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 $\overline{R} \overline{x} = M \overline{R} \overline{x} = \{0\}$, which is the desired contradiction. Observe indeed that $M^c$ and $M^{c + 1}$ are non-zero by assumption, so that both ideals contain $\overline{x}$.

Note. In order to prove of Claim 5, we can use a result from the Matlis's theory of injetive modules over Noetherian rings instead of the Artin-Rees lemma. It goes as follows.

By construction, the ideal $\overline{R}\overline{x} \simeq \overline{R}/M$ is an essential $\overline{R}$-submodule of $\overline{R}$. Therefore the injective hull $E(\overline{R}\overline{x}) \simeq E(\overline{R} / M)$ contains $\overline{R}$. In particular, $1 \in E(\overline{R}\overline{x})$, so that $M^n \cdot 1 = \{0\}$ for some $n \ge 1$ by [Theorem 18.4.v, 3]. Thus $M^n = \{0\}$, which implies that $\overline{R}$ is local, as desired.

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 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.

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 of $R$. Consequently, there is $n \ge 1$ 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.