$\exists x \in R$ and $\exists n \in\mathbb{N}$ such that $x^{n+1} = x^n \implies x^2 = x$

Note that for $k\geq 0$ we have $x^{n+k}=x^n$. Hence, we get

$$ (x^n - x^{n-1})^2 = x^{n-2} (x^{n+2} + x^{n} -2x^{n+1}) =x^{n-2} \cdot 0 =0$$

and thus $x^n= x^{n-1}$.

Let $\ell$ be such that $2^\ell\geq n$. First of all, notice that if we prove that $(x^2 - x)^{2^\ell} = 0$, then the claim will follow. Therefore, we focus in showing that.

We have that $$(x^2 - x)^{2^\ell} = x^{2^\ell}\sum_{r=0}^{2^\ell}\binom{2^\ell}{r}x^r\cdot (-1)^r = \sum_{r=0}^{2^\ell}\binom{2^\ell}{r}x^{2^\ell + r}\cdot (-1)^r = \sum_{r=0}^{2^\ell}\binom{2^\ell}{r}x^{2^\ell}\cdot (-1)^r,$$

where the latter equality follows from the fact that for any $k$ greater than $n$ we have $x^n = x^k$. Now, the last expression can be written as $$x^{2^\ell}\sum_{r=0}^{2^\ell}\binom{2^\ell}{r}\cdot (-1)^r = x^{2^\ell}\cdot(1 - 1)^{2^\ell} = 0.$$

This is apparently an overkill given Severin's nice answer, but it's good to have a different approach as well :)

Note that $x^m=x^n$ for all $m\ge n$.. Then $$\begin{align}(x-x^2)^n&=\sum_{k=0}^n {n\choose k}(-1)^kx^{n+k}\\ &=\sum_{k=0}^n {n\choose k}(-1)^kx^{n}\\&=x^n\sum_{k=0}^n {n\choose k}(-1)^k\\&=x^n(1-1)^2=0\end{align}$$ and by the funny property of $R$, $x=x^2$.