Let $R$ be a ring that has no nonzero nilpotent commutators. If $e\in R$ is an idempotent, then $e\in Z(R)$. Let $R$ be a ring that has no nonzero nilpotent commutators. If $e\in R$ is an idempotent, then $e\in Z(R)$. 
I have a problem with proving this theorem. I don't know how to understand nonzero nilpotent commutators. 
 A: This appears to be a misstatement of the following theorem:

If $R$ is a ring with no nonzero nilpotent commutators, then all idempotents are central.

The reason is that if $e$ is idempotent, then $[e,er]^2=0$ and $[e,re]^2=0$. The condition that commutators which are nilpotent are zero forces $[e,er]=[e,re]=0$, and if you expand them this says that $er=ere=re$, showing that $e$ is central.

Edit: The OP was consequently edited to have the correctly stated problem.
A: Let $e$ be an idempotent and $x$ be any element of the ring. Calculation shows that $[e,ex]^2 = 0$. But you have no non-zero nilpotent commutator. This meanns that you must have $[e,ex] =0$ which implies that $ex = exe$. In a similar way you must have $[e,xe] = 0$ which means that $xe = exe$. comparing with previous $ex = exe$ we get $ex=xe$. Thus $e \in Z(R)$.   
A: Theorem: Let $R$ be a ring that has no nonzero nilpotent commutators. If $e ∈ R$ is an idempotent, then $e ∈ Z(R)$.
Proof: Let $e^2=e$ be an idempotent, and $y\in R$.  Then $(ey−eye)^2 =0=(ye−eye)^2$ for all $y ∈ R$. Just expand and compute. But $ey-eye=eey-eye=[e,ey]$ is a commutator, as well as $ye − eye = [ye,e]$.
Since there are no nilpotent commutators except for zero, we must have
$ey−eye=ye−eye=0$ for all $y∈R$. Thus $ey=ye$ and $e∈Z(R)$.
A: You made a mistake. The right assertion sounds as follows:
Let $R$ be a ring that has no nonzero nilpotent commutators. If $e \in R$
is an idempotent, then $e \in Z(R)$.
This is Theorem 8 from
http://archive.maths.nuim.ie/staff/sbuckley/Papers/bm_variations.pdf
