I am interested in to know how to find integer solutions $(x,y)$, with $x,y\geq 1$ of $$2^{\frac{yp-2}{(p-2)!}}=px+1,$$ where $p$ is a fixed prime satisfying $p\equiv 1\text{ mod }4$. You can read, if you want, the motivation of next questions below.
Question 1. Can you provide me a simple example of solution $(x,y)$, where $x\geq 1$ and $y\geq 1$ are integers of $$2^{\frac{5y-2}{6}}=5x+1?$$ Preferably I would like to know your reasoning to find such simple example.
If there are no mistakes I've discard the first few cases. Additionally I propose another similar problem, for a different prime number.
Question 2. Can you provide me a simple example of solution $(x,y)$, where $x\geq 1$ and $y\geq 1$ are integers of $$2^{\frac{13y-2}{11!}}=13x+1?$$ Thanks you so much.
Motivation. We assume that there exists an odd perfect number $n=p^{4\lambda+1}m^2$, here $p\equiv 1\text{ mod }4$ is prime satisfying $(p,m)=1$. Since $n$ is perfect one has $$2(p-1)=\frac{p^{4\lambda+2}-1}{n}\sigma(m^2):=Q.\tag{1}$$ Previous identity combined with Fermat's little theorem implies that there exist a positive integer $x\geq 1$ such that $$2^Q=xp+1.\tag{2}$$ On the other hand we multiply the identity $(1)$ by $(p-2)!$ to get $$2(p-1)!=Q(p-2)!,$$ then Wilson-Lagrange theorem implies that there exists a positive integer $y\geq 1$ such that $$Q(p-2)!=yp-2.\tag{3}$$ Then for a fixed prime $p$, and we're interested in primes of the form $p\equiv 1\text{ mod }4$, one has from $(2)$ and $(3)$ that there exist positive integers $x,y\geq 1$ such that $$2^{(yp-2)/(p-2)!}=px+1.$$