Romanian Mathematical Olympiad

Since $0$ is nilpotent, we will assume in the sequel that $x \ne 0$.
Given a positive integer $k < n$, notice that $m+p$ is divisible by $k$ for some positive integer $p < n$; write $m + p = kl$. Since $1 - x^{m+p}$ is invertible, and $1 - x^{m+p} = (1 - x^k)(1 + x^k + \dots + x^{k(\ell-1)}) = (1 + x^k + \dots + x^{k(\ell-1)})(1 - x^k)$, it follows that $1 - x^k$ is invertible.

Next, write $1 - x^k = (1-x)(1+x+\dots+x^{k-1}) = (1+x+\dots+x^{k-1})(1-x)$, where we agree that $x^0 = 1$, to deduce that $x_k = 1 + x + \dots + x^{k-1}$ is invertible for each positive integer $k < n$ and $1-x$ is invertible.

If the $x_k$ were pairwise distinct, then by a cardinality argument $R$ would be an $n$-element skew field, so $x^{n-1} = 1$, that is $(1-x)x_{n-1} = 0$, which is impossible since $1-x$ and $x_{n-1}$ are invertible.

Consequently, $x_p = x_q$ for some indices $p < q$, so $0 = x_q - x_p = x^p x_{q-p}$ which implies $x^p = 0$, since $x_{q-p}$ is invertible.