Say that a $k$-ring is a ring in which $x^k=x$ for all $x$, and write $m\trianglelefteq n$ iff every $m$-ring is an $n$-ring. It's not hard to show (see the end of this answer) that $\trianglelefteq$ is computable, and I think in terms of time complexity I can get it down to $O(n\cdot 2^{(2^m)})$; I'd like to know if we can do better.
To see that $\trianglelefteq$ is decidable, first note that by the completeness theorem for first-order, or even equational, logic) $\trianglelefteq$ is c.e., so it will be enough to show that $\not\trianglelefteq$ is c.e. as well. This in turn will follow from an appropriate finite model property: I claim that if $m\not\trianglelefteq n$ then there is a finite $m$-ring which is not an $n$-ring. To see this, suppose $\mathcal{R}$ is an $m$-ring which is not an $n$-ring. Let $\alpha\in\mathcal{R}$ with $\alpha^n-\alpha\not=0$, and let $\mathcal{S}$ be the subring of $\mathcal{R}$ generated by $\alpha$. This subring is a commutative one-element-generated $m$-ring, and so is in fact finite; moreover, since $\alpha\in\mathcal{S}$ we have that $\alpha$ is not an $n$-ring. And so we're done.
Now looking towards complexity, we can bound the size of a putative finite counterexample to $a\trianglelefteq b$ (so that the appeal to the completeness theorem at the outset is unnecessary after all). Every $m$-ring has characteristic at most $2^m-2$. Since the ring $\mathcal{S}$ is a one-element-generated $a$-ring, it has at most $(2^m-2)(m+1)$-many elements. There are doubly-exponentially-many-in-$m$ such rings, and checking whether a ring of size $\approx 2^m$ is an $n$-ring takes time roughly $2^m\cdot n$ (I think? there might be some subtleties around how the ring is presented here), so at a glance the whole process takes time $O(n2^{(2^m)})$.
