$|2^x-3^y|=1$ has only three natural pairs as solutions Consider the equation $$|2^x-3^y|=1$$ in the unknowns $x \in \mathbb{N}$ and $y \in \mathbb{N}$. Is it possible to prove that the only solutions are $(1,1)$, $(2,1)$ and $(3,2)$?

Assume that $x>3$ and $y>2$.
$3^2=1\pmod{8}$. Since $3\not=-1\pmod{8}$ we know that $3^y\not=-1\pmod{8}$. Thus, if $|2^x-3^y|=1$, we must have $2^x-3^y=-1$ and $y$ must be even. Thus, we get that 
$$
2^x=(3^{y/2}-1)(3^{y/2}+1)\tag{1}
$$
The only factors of a power of $2$ are other powers of $2$, and the only powers of two that differ by $2$ are $2$ and $4$, but since $y>2$, we have $3^{y/2}-1>2$ and $3^{y/2}+1>4$.
Therefore, there are no $x>3$ and $y>2$ so that $|2^x-3^y|=1$.
