If $R$ is zero-dimensional, then $\mathrm{Spec}(R)$ is Hausdorff and totally disconnected

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.)

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.)