For a commutative ring $R $ with identity we know that if the prime spectrum, the set of all prime ideal with Zariski topology, is noetherian then max spectrum, the set of all maximal ideal, is also noetherian as a subspace. I am looking for a equivalent condition for $R $ under which the conversation be true.
Note that a topological space is called noetherian if it satisfies the descending chain condition for closed subsets
If we denote $Spm(R)$ the maximal spectrum of $R$, then $Spm(R)$ is a Noetherian topological space (with the topology induces from $Spec(R)$ with the Zariski topology) at least if:
As we know, if $\mathbb{K}$ is a field and $x$ is an indeterminate on $\mathbb{K}$, then $\mathbb{K}[x]$ is a Noetherian non (semi)local ring.
But there exist semilocal non Noetherian rings, for example: if $\mathbb{K}$ is a field and $\{x_1,\dots,x_n,\dots\}_{n\in\mathbb{N}}$ is a infinite many countable family of indeterminates on $\mathbb{K}$, then $R=\mathbb{K}[x_n\mid n\in\mathbb{N}]_{\displaystyle/(x_n\mid n\in\mathbb{N})^2}=\mathbb{K}[\epsilon_n\mid n\in\mathbb{N}]$ is a local ring with maximal ideal $(\epsilon_n\mid n\in\mathbb{N})$; trivially $R$ is a local non Noetherian ring and moreover $Spec(R)=Spm(R)=\{(\epsilon_n\mid n\in\mathbb{N})\}$ is (obviously) a Noetherian topological space.
Without change the names, you can consider the ring $\prod_1^mR=S$, $S$ is a semilocal non Noetherian ring and again: $Spec(S)=Spm(S)$ is a Noetherian topological space with $m$ elements.
From all this, one can draw up a list of families of rings $R$ such that $Spm(R)$ are Noetherian topological spaces.