Associative ring with identity, inverses, divisors of zero and Artinianity

How to prove the following?

$R$ is an associative ring with identity. $R$ contains element $r$. The element is not invertible on the right and is not a left divisor of zero. Then the ring $R$ cannot be Artinian on the right.

Suppose $R$ is right Artinian. Then $rR \supset r^2 R \supset r^3 R \supset \ldots $ is a descending sequence of right ideals. Thus for some $n$, $r^n R = r^{n+1} R$. Use this fact to show that either $r$ is right invertible, or $r$ is a left divisor of zero.

Since $R$ has an identity 1, from $r^nR=r^{n+1}R$ it follows that $r^n1=r^{n+1}s$, or rewritten: $r^n=r^n(rs)$. Since $r$ is not a left divisor of $0$, $r^n\neq0$ and $r^{n+1}s\neq 0$, so it follows that $1=rs$. Then $s$ is a right inverse of $r$, but it contradicts with the statement. Hence $R$ is not Artinian.

This is a generalization of the commutative ring theorem that every Artinian integral domain is a field (with essentially the same proof).