Using the order-reversing correspondence between radical ideals of $R$ and closed subsets of $\operatorname{Spec}(R)$, acc on radical ideals is equivalent to dcc on closed subsets of $\operatorname{Spec}(R)$, which is by definition equivalent to $\operatorname{Spec}(R)$ being a Noetherian topological space.
Noetherian spaces have finitely many irreducible components, this translates to $R$ having only finitely many minimal prime ideals.
If we assume for a moment that $\operatorname{dim}(R) = 0$, then every prime ideal of $R$ is both maximal and minimal, so in particular, there are only finitely many and we have by the Chinese remainder theorem$$R/J(R) = R/\operatorname{nil}(R)\cong \displaystyle \prod_{\mathfrak{p} \in \operatorname{Spec}(R)}R/\mathfrak{p}$$ where each $R/\mathfrak{p}$ is a field.
This shows that $R/J(R)$ is a (commutative) semisimple ring, because it is a finite product of fields. Since we assume that $J(R)=\operatorname{nil}(R)$ is nilpotent, this means that $R$ is a semiprimary ring. (This notion is more prevalent in noncommutative algebra, it means that $J(R)$ is nilpotent and $R/J(R)$ is semisimple.)
Conversely, if we have any semiprimary commutative ring, then by the nilpotency of $J(R)$, we have $J(R)=\operatorname{nil}(R)$ (which is thus nilpotent) and $R/J(R)$ is a finite product of fields, thus $\operatorname{Spec}(R) \cong \operatorname{Spec}(R/J(R))$ (the isomorphism is as topological spaces) is finite and thus $R$ satisfies the acc on radical ideals. (And furthermore $R$ is $0$-dimensional.)
So we can say that the $0$-dimensional rings which satisfy your requirements are precisely the commutative semiprimary rings. These need not be Noetherian. A commutative counterexample is $k[x_1, x_2, \dots]/I$ where $I$ is generated by $x_ix_j$ for $i,j \in \Bbb N$ ($i=j$ is explicitly allowed) for a field $k$.
But there's an interesting theorem regarding Noetherian modules over semiprimary rings, see Hopkins-Levitzki theorem.
Edit: I'll call rings with the acc on radical ideals Spec-Noetherian for now, because that's the characterization I'll be using in the following a lot.
These are some (relatively) elementary properties of Noetherian spaces:
- Noetherian spaces have finitely many irreducible components
- Subspaces of Noetherian spaces are Noetherian
- If $X=\bigcup_{i=1}^n X_i$ is a finite union of Noetherian spaces, then $X$ is Noetherian
- $X$ is Noetherian iff every open subspace is quasi-compact
(In particular if $R=\prod_{i=1}^n R_i$ is a finite product of rings, then $R$ is Spec-Noetherian iff each $R_i$ is Spec-Noetherian.)
Using this, we can give another characterization of Spec-Noetherian rings.
Open subsets of $X=\operatorname{Spec}(R)$ are of the form $X \setminus V(\mathfrak{a})$ where $\mathfrak{a}$ is a radical ideal. Since $D(f)=X\setminus V(f)$ where $f$ ranges over $R$ is a basis of $X$ consisting of compact open subspaces, the compact open subsets $U$ of $X$ are precisely the sets which may be written as a finite union $U=\bigcup_{i=1}^n D(f_i)$, where $f_i \in R$. (Cf. Atiyah-Macdonald exercise 1.7)
So $R$ is Spec-Noetherian iff for every radical ideal $\mathfrak{a} \subset R$, we can write $X \setminus V(\mathfrak{a}) = \bigcup_{i=1}^n D(f_i) = \bigcup_{i=1}^n (X \setminus V(f_i)) = X \setminus \bigcap_{i=1}^n V(f_i) = X \setminus V(f_1, \dots ,f_n)$, so $V(\mathfrak{a})=V(f_1, \dots f_n)$ which is equivalent to $\mathfrak{a}=\sqrt{(f_1, \dots, f_n)}$
So we get the following characterization of Spec-Noetherian rings, very similar to the characterizations of Noetherian rings.
Lemma A ring $R$ satisfies the acc on radical ideals iff every radical ideal is the radical of a finitely generated ideal.
For Noetherian rings, it is enough to know that every prime ideal is finitely generated. We can show the analogous result for Spec-Noetherian rings. For convenience, I'll use results about Oka families, to show this, since some part of this kind of proofs is always the same. (See here for an introduction))
Let $R$ be a commutative ring and let $\mathcal F$ be the family of ideals $I$ in $R$ such that $\sqrt{I}=\sqrt{(a_1, \dots, a_n)}$ for some $a_1, \dots, a_n \in R$. I claim that $\mathcal F$ is Oka family.
We will use the following fact: If $I,J$ are ideals and $a\in R$, then $\sqrt{IJ}=\sqrt{I} \cap \sqrt{J}$
Suppose $I$ is an ideal of $R$ such that $(I:(a)), I+(a) \in \mathcal F$.
We get $(I+(a)) \cdot (I:(a)) = I(I:(a))+(a)(I:(a)) \subset I$
Thus we get $\sqrt{(I+(a))\cdot (I:(a))} \subset \sqrt{I}$.
On the other hand $I \subset I+(a)$ and $I \subset (I:(a))$, so $I^2 \subset (I+(a))\cdot (I:(a))$, thus $\sqrt{I}=\sqrt{I^2} \subset \sqrt{(I+(a)) \cdot (I:(a))}$
So if we suppose $\sqrt{I+(a)}=\sqrt{(a_1, \dots, a_n)}$ and $\sqrt{(I:(a))}=\sqrt{(b_1, \dots, b_m)}$, we can conclude
$\sqrt{I} = \sqrt{(I+(a))\cdot (I:(a)} = \sqrt{(a_1, \dots, a_n)} \cap \sqrt{(b_1, \dots, b_m)} ) = \sqrt{(a_1, \dots, a_n) \cdot (b_1, \dots, b_m)}=\sqrt{(a_1b_1, \dots , a_nb_m)}$
So $I \in \mathcal F$.
Thus $\mathcal F$ is an Oka family and we get the result.
Lemma An ideal maximal with the property that its radical is not the radical of a finitely generated ideal is prime.
Suppose that not every radical ideal of $R$ is the radical of a finitely generated ideal. Then we can use Zorn's lemma to show that the set of radicals which are not the radical ideal of a finitely generated ideal is prime.
Indeed, if we have an ascending chain $I_1 \subset I_2 \subset \dots$ of radical ideals which are not the radical of a finitely generated ideal, then $I= \bigcup_{k=1}^\infty I_k$ is a radical ideal. If we had $I=\sqrt{(a_1, \dots a_n)}$, then every $a_i$ would be contained in some $I_k$, thus as we have an ascending chain, $I_k$ would contain all the $a_i$ for $k$ large enough. But then $I_k=I=\sqrt{(a_1, \dots , a_n)}$ which is impossible. Thus be conclude by Zorn's lemma.
Combing this with the previous result, we get.
Lemma A ring satisfies the acc on radical ideals iff every prime ideal is the radical of a finitely generated ideal.
If we use the fact that $\operatorname{Spec}(R) = \bigcup_{\mathfrak{p} \in \operatorname{SpecMin}(R)} V(\mathfrak{p})$ is the decomposition into irreducible components, where $\operatorname{SpecMin}(R)$ is the set of minimal prime ideals, we get the following:
Lemma A ring $R$ is Spec-Noetherian, iff $R$ has finitely many minimal prime ideals and for every minimal prime ideal $\mathfrak{p}$ $R/\mathfrak{p}$ is Spec-Noetherian.
But $R/\mathfrak{p}$ is an integral domain so we have reduced the problem in some sense to integral domains.
There's a somewhat "dual" criterion that is only sufficient in general
Lemma If $R$ has only finitely many maximal ideals, then $R$ is Spec-Noetherian iff $R_{\mathfrak{m}}$ is Spec-Noetherian for each maximal ideal $\mathfrak{m}$
Proof: $\operatorname{Spec}(R) = \bigcup_{\mathfrak{m} \in \operatorname{SpecMax}(R)} \operatorname{Spec}(R_{\mathfrak{m}})$, so if this union is finite, we can apply the property of Noetherian spaces.
To see how this fails for rings with infinitely many maximal ideals, consider non-Noetherian von-Neumann-regular ring, e.g. an infinite product of fields.
For integral domains, I have no idea how to completely characterize those that are Spec-Noetherian, but here are some ideas for a special class of integral domains:
If $R$ is a valuation ring, then the ideals of $R$ are totally ordered by inclusion, so every radical ideal, being the intersection of a totally ordered chain of prime ideals is prime. Furthermore, since all the prime ideals form one chain, we get this easy lemma:
If $R$ is a valuation ring, then the following are equivalent:
- $R$ is Spec-Noetherian
- The Krull dimension of $R$ is finite
- $\operatorname{Spec}(R)$ is finite
- $\forall \mathfrak{p} \in \operatorname{Spec}(R)$ $\mathfrak{p}=\sqrt{(a)}$ for some $a \in R$
(The last point follows from the previous lemma on radical ideals, because every finitely generated ideal in a valuation ring is prinicpal.)
This makes it easy to construct examples of non-Noetherian Spec-Noetherian local integral domains of any given finite (non-zero) Krull dimension, because we can control the Krull dimension of $R$ by choosing the appropriate value group. (See for example here) And we can also construct local integral domains which are not Spec-Noetherian this way.
