If $R$ is zero-dimensional, then $\mathrm{Spec}(R)$ is Hausdorff and totally disconnected 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 ?
 A: If every prime is maximal, then $R/\mathord{\operatorname{nil}(R)}$ is von Neumann regular. 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.)
A: To say that every prime ideal is maximal is to say that each point of Spec $R$ is closed.   One can then prove purely topologically that $X$ is Hausdorff (using
the basic topological properties of Spec's).  (See this question and its answers.)
A: 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 )
