If we suppose, that a perfect odd number exists, can we determine whether infinite many exist? 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.
 A: 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.
A: This answer attempts to prove your conjecture that if there is an odd perfect number, then there are infinitely many of them.
Briefly, the argument makes use of the following lemma:

Lemma If $q^k n^2$ is an odd perfect number with special/Euler prime $q$, then the biconditional
  $$\frac{q^2}{n}+\frac{n}{q^2} \text{ is bounded from above } \iff \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} \text{ is bounded from above}$$
  holds.

Details are in the following question:
If $q^k n^2$ is an odd perfect number with Euler prime $q$, which of the following relationships between $q^2$ and $n$ hold?
