$2^m+3^n$ is a square Determine all pairs ($m,n$) of positive integers such that  $2^m+3^n$ is a square.
One such pair is ($4,2$) and I think that there are no other solutions. Am I correct? If I am can somebody please give me a hint to prove it?

If $m=1$ , then we have $2+3^n$, which cannot be a square because the residue modulo $3$ is $2$.
If $m>1$, then $n$ cannot be odd because the residue of $2^m+3^n$ modulo $4$ would be $3$. So, $2^m$ and $3^n$ must both be squares. So, we need a pythagorean triple $(u,v,w)$ , where $u$ is a power of $3$ and $v$ is a power of $2$.
The general solution is $u=m^2-n^2$ , $v=2mn$ , so $mn$ must be a power of $2$, so $m$ and $n$ must both be a power of $2$, while $(m-n)(m+n)$ must be a power of $3$. If $m>1$ and $n>1$, then $(m-n)(m+n)$ is even. $m=1$ is also impossible, so we must have $n=1$. So, $(2^k-1)(2^k+1)$ must be a powe of $3$
But $2^k+1$ is never a power of $3$ for $k>3$ because of the proven catalan-conjecture. If $k=2$, then $2^k+1=5$, if $k=3$, then $2^k+1=9=3^2$, but $2^k-1=7$. So, we must have $k=1$ and therefore $m=2$ and thereofore $u=3$ , $v=4$, corresponding with the only solution $2^4+3^2=4^2+3^2=5^2$
