MHB Can a left Noetherian ring have $xy = 1$ without also having $yx = 1$?

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Here's this week's problem!Problem. Prove that if $A$ is a left Noetherian ring, then $xy = 1$ implies $yx = 1$, for every pair $x,y\in A$.
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No one answered this week's problem. You can find my solution below.

Spoiler

Let $x,y\in A$ with $xy = 1$. Since $A$ is left Noetherian, the infinite chain $\operatorname{Ann}(y) \subseteq \operatorname{Ann}(y^2) \subseteq \cdots$ stabilizes, so there exists a positive integer $n$ for which $\operatorname{Ann}(y^n) = \operatorname{Ann}(y^{n+1})$. Let $z = yx - 1$. Then $zy = y(xy) - y = y - y = 0$, whence $z\in \operatorname{Ann}(y)$. Since $xy = 1$, $x^n y^n = 1$, so $z = zx^n y^n$. Thus $0 = zy = zx^ny^{n+1}$, which implies $zx^n\in \operatorname{Ann}(y^{n+1})$. Then $zx^n \in \operatorname{Ann}(y^n)$, i.e., $zx^ny^n = 0$. Since $x^n y^n = 1$, $z = 0$. Hence, $yx = 1$.