My proof: Suppose that $R$ is a division ring. Since $0a=a0=0$, then $0^2 =0$. If $a \neq 0$ and $a^2 = a$, then the inverse $a^{-1}$ of $a$ exists. So, $a^{-1} (a^2)=a^{-1} (a)$. This implies that $a=1$.
Here's another proof that I found: If $a^2=a$, then $a^2 -a = a(a-1) =0$. If $a \neq 0$, then $a^{-1}$ exists in $R$ and we have $a-1 = (a^{-1}a)(a-1)=a^{-1}[a(a-1)]=a^{-1}0 =0$, so $a-1=0$ and $a=1$. Thus $0$ and $1$ are the only two idempotent elements in a division ring.
I think my proof is more elegant, but are some steps of it incorrect that must require me to change it the way that the second proof is written?