Showing that the only idempotents in $R$ are zero and one I have the following question that I have to solve however I cannot achieve.

Let $R$ be a ring with $1$ and suppose $R$ has no zero divisors. Show that the only idempotents in $R$ are $0$ and $1$.

Please help me to solve this question.
Your prompt reply will be greatly appreciated.
Thanks in advance.
 A: Suppose $e$ is idempotent. Then $e^2=e$, so $e^2-e=0$. Trying factoring and see what happens, using the fact that there are no zero divisors.
A: If $a^2 = a$ (the definition of idempotence), we have $a(a - 1) = 0$. Since $R$ has no zero-divisors, either $a=0$ or $a = 1$.
A: Hint $\ $ The idempotents $\,0\,$ and $\,1\,$ are roots of a nonzero quadratic polynomial $\,f\in R[x],\,$ which has at most two roots, since $\,R\,$ is a domain. Indeed, if  $\,f\,$ has distinct roots $\,r,s\,$ then, by the Bifactor Theorem we deduce $\,f(x) = a(x-r)(x-s),\ $ so if  $\ 0 = f(t) = a(t-r)(t-s)\ $ then $\,a\ne 0\,\color{#c00}{\Rightarrow}\,(t-r)(t-s)=0\,\color{#c00}\Rightarrow\,t-r=0\,$ or $\,t-s=0,\,$ therefore  $\,t=r\,$ or $\,t = s,\,$ where the $\color{#c00}\Rightarrow$'s follow from $\,R\,$ is a domain, i.e. $\,R\,$  has no zero-divisors $\ne 0$.
Remark $\ $ Since at least one reader seems to have misinterpreted the above, let me clarify that the above more general proof is intended for commutative rings (I explicitly say "$R$ is a domain" and domains are commutative by standard conventions). However, even if the OP intends $R$ to be possibly noncommutative,  the latter half of the proof still yields the desired inference when applied to $\,f(x) = x(x-1).$
