A ring $R$ is said to be
- reduced if the only nilpotent in $R$ is $0$.
- Dedekind finite if all one-sided units in $R$ ($ab=1$) are two-sided units ($ab=ba=1$).
- stably finite if the matrix ring $M_n(R)$ is Dedekind finite for all positive integers $n$.
- a ring satisfying Klein's nilpotence condition if for any positive integer $n$ and any nilpotent matrix $A \in M_n(R)$, $A^n=0$.
Any reduced or stably finite ring is Dedekind finite. But it is known that there exist domains (which are automatically reduced and Dedekind finite) that are not stably finite (see MathOverflow).
Also, any ring satisfying Klein's nilpotence condition is both reduced and stably finite.
But is there a stably finite reduced ring that does not satisfy Klein's nilpotence condition?
Let $K$ be a field and $R$ be the quotient of the free $K$-algebra on $4$ generators $w, x, y, z$ by the two-sided ideal generated by the entries of the matrix $\begin{pmatrix} w&x\\y&z \end{pmatrix}^3$.
Then, clearly, $R$ does not satisfy Klein's nilpotence condition. But is $R$ reduced and stably finite?
Any commutative reduced ring can be embedded in the product of the fields of fractions of the quotient rings by its prime ideals, and fields are known to satisfy Klein's nilpotence condition, so there are no counterexamples in the commutative case.