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Let $R$ be a ring with $1$. Is it true that $R$ has IBN $\iff$ $R^n \ncong R^{n+1}$ for all $n \in \mathbb{N}$?

Per definition $R$ has IBN (invariant basis number) if $R^{m} \cong R^{n}$ as left $R$-modules implies $n = m$.

I don't think that the above is true but I have no counterexample. The example I know is only $R = \text{End}_k(V)$ for some infinite-dimensional $k$-vector space $R$. In that case it is $R^2 \cong R$.

But does someone have a preferably simple counterexample to the above?

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    $\begingroup$ No. See the remarks at mathoverflow.net/q/36644 Leavitt produced examples of rings of type $(h,k)$ for any $h, k\geq 1$. A ring of type $(1,2)$ is a counterexample. $\endgroup$
    – Keith Kearnes
    Commented Jun 17 at 15:52

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