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.

I am assuming that $x$ and $y$ are supposed to be integers.
The claim follows from the fact that if $x>3$, then the order of the residue class of $3$ in the group $\mathbb{Z}_{2^x}^*$ (sometimes denoted by $U_{2^x}$)
is $2^{x-2}$. In other words: for $3^y-1$ to be divisible by $2^x$ the exponent $y$ has to be a multiple of $2^{x-2}$. There are several proofs for this fact in this site.
But when $y\ge 2^{x-2}$, then it shouldn't be too hard to see that $$3^y\ge3^{2^{x-2}}>2^x.$$
