This does not answer the general question, but a simple special case, which I find cute:
Suppose $R$ is a finite dimensional algebra over a field of characteristic zero. Pick $x$, $y\in R$. Since $[x,y]$ is a commutator, its action on every finite dimensional $R$-module is of trace zero; on the other hand, since $[x,y]$ is by hypothesis idempotent, its trace on a module is the dimension of its image. It follows that $[x,y]$ acts by zero on every module so, because faithful modules do exist, it must be zero.
Later. Consider the non-commutative polynomial $f(x,y)=[x,y]^2-[x,y]$, and define new polynomial $h$ by $$h(x,y)=f(x,y)-f(y,x)$$ A simple computation shows that $h(x,y)=2[x,y]$.
If we plug elements of $R$ into $f$ we get zero, so the same is true of $h$. The above observation means then that $2[x,y]=0$ for all $x$, $y\in R$. If now $2$ is invertible in $R$, or at least not a divisor of zero, then we see that $R$ is commutative.