$3^x-2^y=1$, $x \in \mathbb{N}$ and $y \in \mathbb{N}$ $3^x-2^y=1$
or
$y=\log_2{\left(3^x-1\right)}$
$x$ and $y$ must be natural numbers.
I know this two solutions:
$x=1$ and $y=1$
$x=2$ and $y=3$
Are there more solutions?
How can I find them?

We look for $y\ge 2$.  Then we must have $3^x\equiv 1\pmod{4}$, so $x$ must be even, say $x=2a$. Now we can rewrite our equation as $3^{2a}-1=2^y$, or equivalently as $(3^a-1)(3^a+1)=2^y$. 
It follows that $3^a-1$ and $3^a+1$ must each be powers of $2$. They differ  by $2$, and the only pair of powers of $2$ that differ by $2$ are $2$ and $4$. That forces $a=1$. So there are no solutions in positive integers other than the ones you listed.
