It is currently not known whether odd perfect numbers (numbers with the property $\sigma(n)=2n$, where $\sigma(n)$ is the sum of divisors of $n$, including $1$ and $n$) exist.
But suppose, a perfect odd number exists.
Do we then know whether there are infinite many perfect odd numbers ?
This could be possible because various necessary conditions are known for an odd number to be perfect. Perhaps they allow to construct arbitary many perfect odd numbers, supposing that an odd perfect number exists.
This is not a direct answer to your question, but it is certainly related.
We do know that $$I(q^k) < \frac{5}{4} < \frac{3}{2} \leq I(2^{p-1})$$ where $N=q^k n^2$ is an odd perfect number in Eulerian form and $M=(2^p - 1){2^{p-1}}$ is an even perfect number in Euclidean form.
Because of a property satisfied by the abundancy index $I(x)=\sigma(x)/x$ at prime powers, it follows that $$2^{p-1} < q^k.$$
(NOTE THAT THE REASONING IS UNSOUND IN THE LAST PARAGRAPH, AS IT ONLY FOLLOWS FROM $I(q^k) < I(2^{p-1})$ THAT $2 < q$. WHAT FOLLOWS IS THEREFORE INCONCLUSIVE, AT THIS POINT.)
It follows that, since $q^k < n^2$ [Dris, 2012], if $N$ is bounded, then $M$ is also bounded.
Thus, if there exists an odd perfect number, we would infer that there are only finitely many even perfect numbers.