I think only pairs are $(0,1)$ and $(3,3)$ so answer is $9$.

May be you have used trial and error method. But can you give formal argument ?

$n^2-1=2^m$ so $n-1=2^a,n+1=2^b,a+b=m$ $2^{b-a}=\frac{n+1}{n-1}=1+\frac{2}{n-1}$ so $n$ can be only $2,3$ but for $n=2$ there are not solutions, so $(m,n)=(3,3)$

RagvaloD wrote: $n^2-1=2^m$ so $n-1=2^a,n+1=2^b,a+b=m$ $2^{b-a}=\frac{n+1}{n-1}=1+\frac{2}{n-1}$ so $n$ can be only $2,3$ but for $n=2$ there are not solutions, so $(m,n)=(3,3)$ But in the term $2^{b-a}$ when $a>b$ then what happen?

If $a>b$ then $2^{b-a}<1$, but $1+\frac{2}{n-1}$ is clearly greater than $1$, so we have no solutions.

By Mihaeliscu's Theorem (I don't know how to spell it ), we know that the only consecutive powers are $(8, 9)$ and $(1, 2)$. Clearly we can't have $m = 0$, otherwise $n$ wouldn't be an integer. So it would be $(8, 9)$ the pair. This would imply that $m = 3$ and $n = 3$ hence $mn = \boxed{9}$.

By Catalan' theorem: $\implies 2^m=8 ; n^2=9$ $\implies m=3 ; n=3$ $\implies mn=9$