= Solution
Reflexivity follows by taking the two exponents equal to $1$, and symmetry is built into the two divisibility conditions. For transitivity, suppose $m\mid n^a$, $n\mid m^b$, $n\mid \ell^c$ and $\ell\mid n^d$. Then $m\mid\ell^{ac}$ and $\ell\mid m^{bd}$, with positive exponents. Thus the relation is an <equivalence relation>.
Its <equivalence classes> have a useful description by <prime factorization>. Let $S(n)$ be the finite set of <prime factors> of $n$, with $S(1)=\varnothing$. If $m\mid n^a$, every <prime factor> of $m$ divides $n$; the reverse divisibility gives $S(n)\subseteq S(m)$. Conversely, if $S(m)=S(n)$, write $m=\prod_{p\in S}p^{\alpha_p}$ and $n=\prod_{p\in S}p^{\beta_p}$, with all exponents positive. Choosing $a\ge\max_p\lceil\alpha_p/\beta_p\rceil$ and $b\ge\max_p\lceil\beta_p/\alpha_p\rceil$ gives the required divisibilities. The empty case is exactly $m=n=1$.
Therefore the <prime-support equivalence relation> is
$$
\boxed{m\sim n\iff S(m)=S(n)\iff\operatorname{rad}(m)=\operatorname{rad}(n),}
$$
where the <radical of an integer> is the product of its distinct <prime factors>. The class with empty support is $\{1\}$. For every nonempty finite support $S$, all positive exponent choices give one class, and varying just one exponent produces infinitely many different <integers> in it.
There are infinitely many classes because each <prime number> gives a different singleton support. For completeness, if there were only finitely many <prime numbers> $p_1,\ldots,p_r$, a <prime factor> of $p_1\cdots p_r+1$ would differ from them all. \b[There are infinitely many classes; the unique finite class is $\{1\}$.]
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