One minus a nilpotent element is a unit ($(1-z{)}^{-1}=1+z+\cdots +z^{m-1}$)

If $z$ is a nilpotent element of a ring, choose $m\ge 1$ with $z^m=0$. The finite geometric series identity givesand the factors commute, so $1-z$ is a unit. In a commutative ring, every multiple of a nilpotent element is again nilpotent. Consequently an equality $x=a{x}^2$ with $x$ nilpotent forces $x=0$: rewrite it as $(1-ax)x=0$ and multiply by the inverse of $1-ax$.