Are there any positive integer solutions to $2^n-3^m=1$ with $n,m>2$ ?


By way of justifying the question, I've found lots of info on what happens when $m=n$ (mostly FLT variations, Darmon + Merel,...), but I don't really know where to look for $m\not=n$.


Also it's pretty obvious that you can't have solutions to similar equations, e.g., $2^n-3^m=2$. There are no solutions for $n,m<1000$ aside from $n=2, m=3$. It seems pretty likely to me that it should happen for some large numbers at some point though. 


Are there any theorems I don't know about regarding primes $p,q$ and $p^n-q^m=k$, $k \in \mathbb{N}$ that might rule out a solution or help me find one?



Are there any solutions to $2^n-3^m=1$?
Here is the proof of Gersonides [Levi ben Gershon] (1343) for $2^n-3^m=1$. It uses nothing more that arithmetic modulo $8$.


Case I: $m$ is even. Then $3^m$ is 1 mod 4, so $2^n$ is 2 mod 4, implying $n=1$ and $m=0$.


Case II: $m$ is odd. Then $3^m$ is 3 mod 8, so $2^n$ is 4 mod 8, implying $n=2$ and $m=1$.


The alternative equation $3^m-2^n=1$ follows similarly when $m$ is odd, but is a bit more tricky when $m$ is even (hint, factor $2^n=3^m-1=(3^{m/2}+1)(3^{m/2}-1)$ and argue from there).


