$r$ is not nilpotent, $r-r^2$ is nilpotent, then the ring has a non-zero idempotent 
Assume that $R$ is a ring and $r-r^2$ is nilpotent for an element $r\in R$. If $r$ is not nilpotent, then $R$ has a nonzero idempotent.

 A: Let $(r-r^2)^n=0$. Then we can conclude that $r^n=r^{n+1}f(r)$. It is not hard to see that $(r^{n}f(r)^n)^2=r^{2n}f(r)^{2n}$. Now, we compute $r^n$. Note that $r$ and $f(r)$ commuate
$$r^n=r^{n+1}f(r)=rf(r)r^n=rf(r)r^{n+1}f(r)=r^2f^2(r)r^n=r^{n+2}f^2(r)$$
We can repeat procedure and we can find $r^n=r^{2n}f^n(r)$. Thus we have $r^{2n}f^{2n}(r)=r^{2n}f^n(r)f^n(r)=r^nf^n(r)$ and the idempotent element is $r^nf^n(r)$.
A: By considering the subring generated by $r$, we may assume that $R$ is commutative. But then we can give a "geometric" solution as follows.
If $\mathfrak{p} \subseteq R$ is a prime ideal, then we have $r(1-r) \in \mathfrak{p}$, hence $r \in \mathfrak{p}$ or $1-r \in \mathfrak{p}$. Both $r$ and $1-r$ cannot be contained in $\mathfrak{p}$. This shows that $\mathrm{Spec}(R)$ is the disjoint union of the closed subsets $V(r)$ and $V(1-r)$. Since $r$ and $1-r$ are not nilpotent (see my comment), these closed subsets are proper subsets. It follows that $\mathrm{Spec}(R)$ is disconnected and hence has a nontrivial idempotent element.
