The prime spectrum is connected exactly when the ring has no idempotents other than zero and one

Suppose $\mathrm{Spec}(R)$ is connected. If $e^2=e$, then [L2] gives a clopen partition by $D(e)$ and $D(1-e)$. By [L1], one of these clopen subsets is empty. If $D(e)=\mathrm{\varnothing }$, then every prime ideal contains $e$, so [L4] shows that $e\in \mathrm{Nil}(R)$. Thus $e$ is nilpotent, and the idempotent relation $e^2=e$ forces $e=0$. If $D(1-e)=\mathrm{\varnothing }$, then every prime ideal contains $1-e$, so [L4] gives $1-e\in \mathrm{Nil}(R)$. Hence $1-e$ is nilpotent, and $(1-e{)}^2=1-e$ forces $1-e=0$, so $e=1$. Thus there is no nontrivial idempotent.

Suppose conversely that $\mathrm{Spec}(R)$ is disconnected. Then [L1] gives a nonempty proper clopen subset $C$. By [L3], there is an idempotent $e\mathrm{\notin }\{0,1\}$ whose associated clopen subset is $C$. Thus $R$ has a nontrivial idempotent.

Step 1.1 proves $(1)\Rightarrow (2)$ and step 1.2 proves its contrapositive reverse direction. Therefore $\mathrm{Spec}(R)$ is connected exactly when $R$ has no idempotents other than $0$ and $1$.