Are there distinct primes $p,q$ satisfying $pq=(2^r-1)(p+q)-5$?

Question

We let $p\neq q$ be odd prime numbers and $r$ be integer $>2$. Are there such $p,q$ satisfying $pq=(2^r-1)(p+q)-5$?

This is clear from here that,

$q(p-2^r+1)=(2^r-1)p-5$,

and $p(q-2^r+1)=(2^r-1)q-5$.

Thanks.

Answer 1

Over at tomerg's other, closely-related question The form $xy+5=a(x+y)$ and its solutions with $x,y$ prime I found $p=17179929661$, $q=4880269588100161$, $r=34$ is a solution.

Answer 2

Blockquote EDIT: After writing, the original problem was edited to read $p(q - 2^r - 1)...$ and so this is all beautiful speculation on a different problem. Had I checked his derivation, I might have noticed. But I didn't.

Firstly, I wonder - is there a significance to this question? I've no idea. But I put together some scratch work real quick and found a solution - so there's that.

I note firstly that at most one can be even (considering the equation mod 2 gives this). So I thought, what if p were 2?

Then we have $$ 2(q + 2^r - 1) = (2^r - 1)1 - 5$$ $$ 2^{r+1} + 3 = (2^r - 3)q$$ $$ \frac{2^{r+1} + 3}{2^r - 3} = q $$

And if $ r = 2$, we get that $ q = 11$. Although I don't have it yet, I suspect this is the only solution for q for $p = 2$. So there is at least one answer.

Without having a better intuition for the problem (as I don't really know if there's anything special about this equation, if it means something in particular, etc.), I don't see a better method of attack than this sort of play.