How many pairs?

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 $y = 2z$ 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 when $x = 3$ and $y = 2$ , which corresponds to $(n, m) = (1, 1)$ in the original problem.

To be honest: I tried the first week pairs (0,0) and (1,1) and couldn't think of any power of 2 and 3 that had 1 difference, so 2 turned out to be enough to answer the question right. Haven't thought of proving the 2 pairs I found are the only 2 possible pairs.