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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 iswhere 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 $p_1,\dots ,p_r$, a prime factor of $p_1\cdots p_r+1$ would differ from them all. There are infinitely many classes; the unique finite class is $\{1\}$.