Question: I've found the following claim.

If $R$ is a commutative ring, and every prime ideal is maximal, then $\mathrm{Spec}(R)$ with the Zariski topology is Hausdorff and totally disconnected.

Is it true ? Why ?

Answer: 
If every prime is maximal, then [$R/\mathord{\operatorname{nil}(R)}$ is von Neumann regular](https://math.stackexchange.com/a/629307/29335). Since the spectrum of $R$ is homeomorphic to that of $R/\mathord{\operatorname{nil}(R)}$, without loss of generality we can assume $R$ is von Neumann regular.

The thing about von Neumann reuglar rings is that they're full of idempotents, and idempotents disconnect the spectrum. Let $e$ be an idempotent. Notice that given a prime ideal, exactly one of $e$ or $1-e$ is in the prime ideal. This allows you to demonstrate that $V(eR)\cup V((1-e)R)=\operatorname{Spec}(R)$, and $V(eR)\cap V((1-e)R)=\emptyset$. Thus every set of the form $V(eR)$ is clopen.

Now suppose $X\subseteq \operatorname{Spec}(R)$ is a subset with more than one element. Take two of these elements, say $M\_1$ and $M\_2$. Take $a\in M\_1\setminus M\_2$. Since $R$ is VNR, $aR=eR$ where $e$ is an idempotent, and we now have $e\in M\_1\setminus M\_2$. Thus $M\_1\in V(eR)$ and $M\_2\in V((1-e)R)$. Thus these two clopen sets of $\operatorname{Spec}(R)$ disconnect the set $X$.

(Of course, this simultaneously establishes that any two points can be separated with open sets.)
if every prime ideal is maximal, then the krull dimension of R is 0.
Then choose M,N in Spec(R). AS they are maximal, M+N=R. then m+n=1, and D(m) is a basic open set in zariski's topology which contains M, and D(n) is a basic open set which contains N. D(m) and D(n) are disjoint and therefore Spec(R) is Hausdorff. (M different from N )