Are there any integer solutions to $2^x+1=3^y$ for $y>2$? for what values of $ x $ and $ y $ the equality holds 

$2^x+1=3^y$

It is quiet obvious the equality holds for $x=1,y=1$ and $x=3,y=2$.
But further I cannot find why $x$ and $y$ cannot take higher values than this values.

Note that $2\equiv -1$ mod $3$, which means that $x$ must be odd.
If $y$ is even, let $y=2z$ and we have $2^x=(3^z+1)(3^z-1)$ with some minor arranging. The two factors on the right-hand side differ by $2$, and must both be powers of $2$, so they must be $2$ and $4$ and $z=1$ which gives us the solution $x=3, y=2$.
It remains to deal with the case in which $x,y$ are both odd. We note that $3\equiv -1$ mod $4$ so if $x \ge 2$ and $y$ is odd the equation becomes $0+1\equiv -1$ mod $4$. So we must have $x\lt 2$ hence $x=1, y=1$.
