Let $k$ be a field of characteristic $0$. If $m\ge2$, I denote $P_m$ the standard polynomial in $m$ non-commutative indeterminates: $$P_m(X_1,\ldots,X_m)=\sum_{\sigma\in{\frak S}_m}\epsilon(\sigma)X_{\sigma(1)}\cdots X_{\sigma(m)}.$$ We say that a $k$-algebra $A$ is $m$-commutative if $P_m$ is an identity over $A$, that is $$\forall a_1,\ldots,a_m\in A,\qquad P_m(a_1,\ldots,a_m)=0,$$ and $P_{m-1}$ is not. Remark that $2$-commutativity is just commutativity.

Amitsur-Levitzki's Theorem is that $M_n(k)$ is $(2n)$-commutative.

My question is whether there exist (interesting) algebras that are $m$-commutative for an odd $m$. For instance, what would be an example of a $3$-commutative algebra ?

Edit. As mentionned by user49822, $A$ may not be unital, overwise $(2k+1)$-commutativity implies $(2k)$-commutativity by specifying $a_m=1$. Thus there remains the question of whether a non-unital algebra can be $3$-commutative.