Give as much different solutions - 1

Suppose $n$ is even. Then $3^{n} = (4 - 1)^{n} \equiv 1 \pmod{4}.$

So with $3^{n} = 4k + 1$ for some non-negative integer $k$ we would require that

$2^{m} - (4k + 1) = 1 \Longrightarrow 2^{m} = 4k + 2 = 2(2k + 1) \Longrightarrow 2^{m - 1} = 2k + 1.$

The only odd power of $2$ is $2^{0} = 1,$ so if $n$ is even we must have $m = 1, k = 0,$ and thus $n = 0,$ which does not satisfy the condition that $m$ be a positive integer.

Now suppose $n$ is odd. Then $n = 2k + 1$ for some non-negative integer $k,$ and

$3^{n} = 3^{2k + 1} = 3*9^{k} = 3*(8 + 1)^{k} \equiv 3 \pmod{8}.$

So with $3^{n} = 8d + 3$ for some non-negative integer $d$ we would require that

$2^{m} - (8d + 3) = 1 \Longrightarrow 2^{m} = 8d + 4 = 4(2d + 1) \Longrightarrow 2^{m - 2} = 2d + 1.$

Again, as the only odd power of $2$ is $2^{0} = 1,$ if $n$ is odd we must have $m = 2, d = 0,$ and thus $n = 1.$

Having covered all the cases, we thus see that $(m,n) = (2,1)$ is the only solution pair satisfying the given conditions, and so the answer is $\boxed{1}.$

Taking mod 3 tells us that $m$ is even Let $m = 2k$ . We have $(2^k)^2 - 3^n = 1 \implies (2^k + 1)(2^k - 1) = 3^n$ Thus $2^k \pm 1$ must both be powers of three. But the only powers of three that are different by $2$ are $3$ and $1$ . Setting $\begin{aligned} 2^k + 1 & = 3 \\ 2^k - 1 & = 1 \end{aligned}$ Gives us the unique solution of $(2, 1)$ .

$2^m = 3^n + 1$

$2^1 = 3^0 + 1$ {For non-negative integers}

$2^2 = 3^1 + 1$ {For positive integers}

When n>1, $3^n + 1$ can mostly be divided by $2^2$ . The point is not $2^m$ .

This means that any $3^n + 1$ at far which shall not be checked if equals to $2^m$ cannot be true as two equal values must be divisible by $2^m$ of huge m rather than only a maximum of 4 (to an immediate limit using Excel).

Therefore, $2^2 = 3^1 + 1$ is the only case valid for all positive integers of m and n.

Answer: $\boxed{1}$