Simple: The solutions are $n=2,7.$ For $m=0, n=2$ and $m=1$ we don't have any solution. So let $m\ge 2.$ So we have $3\cdot 2^m=n^2-1.$ Then $3\cdot 2^m=(n-1)(n+1).$ Clearly $n$ is odd. Then by gcd stuffs $n-1=3\cdot 2$ and $v_2(n+1)= m-1.$ So $n=7.$