Find all integers $x,y \in \Bbb N$ such that $2^x = 3^y+5$.
I found solutions $(x,y)=(3,1)$ and $(x,y)=(5,3)$ and now I'm trying to prove that there are no solutions for $x \ge 6, y \ge 4$.
I've found little examples of these kind of exponential diophantine equations and they often consider the order of an element of some kind of cyclic behavior w.r.t to some modulo, but I don't know how to approach this. May I have some hints on what should I do?
Edit: I noted that if I subtract $2$ from both sides the equation is as $2^x-2=3^y+3 \iff 2(2^{x-1}-1)=3(3^{y-1}+1)$ this makes $2^{x-1}-1$ a multiple of $3$ and $3^{y-1}+1$ a multiple of $2$. So we have that $$2^{x-1} \equiv 1 \pmod 3 $$ and that $$3^{y-1} \equiv 1 \pmod 2$$ now the order of $2$ modulo $3$ is $2$ and the order of $3$ modulo $2$ is one $1$ that is $x-1$ is a multiple of $2$ and well $y-1$ is a multiple of $1$. The latter doesn't give anything nice, but if $x-1$ is a multiple of $2$ we have that $$x-1=2k \implies 2(2^{x-1}-1)=2(2^{2k}-1)=2(2^2-1)(2^{2(k-1)} +2^{2(k-2)}+ \dots + 2^2+1)$$ i.e the rhs is a multiple of $3$ also?