We find all solutions of $2^{x}=3^{y}-1$ for positive integers $x$ and $y$. If $x=1$, we obtain the solution $x=1, y=1$, which corresponds to $(n, m)=(0,0)$ in the original problem. If $x>1$, consider the equation modulo 4. The left hand side is 0, and the right hand side is $(-1)^{y}-1$, so $y$ is even. Thus we can write $y=2 z$ for some positive integer $z$, and so $2^{x}=(3^{z}-1)(3^{z}+1)$. Thus each of $3^{z}-1$ and $3^{z}+1$ is a power of 2, but they differ by 2, so they must equal 2 and 4 respectively. Therefore, the only other solution is $x=3$ and $y=2$, which corresponds to $(n, m)=(1,1)$ in the original problem.