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).

The equation $p^n-q^m=k$ is a special case of the $S$-unit equation, so it has finitely many solutions by a theorem of Siegel. Using linear forms in logarithms you can get an explicit upper bound for the size of the solutions and, in principle, find all of them. In practice, the bounds may be too big in general, but I am sure $2^n-3^m=1$ has been done (it also follows from Catalan, as Todd has just commented). Check out Baker's book on Transcendence.

It is the content of Catalan conjecture That is to say, that the only solution in the natural numbers of $x^a -y^b= 1$ for $ x,a, y,b > 1$ is $x=3, a=2, y=2, b=3$

it was proved in 2002 by Preda Mihăilescu.

http://en.wikipedia.org/wiki/Catalan's_conjecture

a good refernce for a self contained proof is the book by Rene' Schoof: "Catalan's Conjecture", Universitext, Springer-Verlag, 2008.