Natural numbers only

Notice that $(m-8)$ and $(m-10)$ are factors of a power of $2$ .

Since all factors of a power of $2$ must be powers of $2$ due to $2$ being the only prime factor, $(m-8)$ and $(m-10)$ must be powers of $2$ if $m-8$ and $m-10$ are assumed positive which occurs for values greater than ten.

Note that $(m-8)$ and $(m-10)$ also differ by $2$ , The only $2$ powers of $2$ that differ by $2$ are $2$ and $4$ , so the answers are whenever one of $(m-8)$ and $(m-10)$ is $2$ and the other is $4$ or whenever one of . $(8-m)$ and $(10-m)$ is $2$ and the other is $4$ . This is due to being able to rewrite the equation as $(8-m)(10-m) = {2}^{n}$ which would be the case where $m-8$ and $m-10$ are assumed negative which would occur for values less than 8.

This happens at $m = 12$ or $m = 6$ . Therefore, there are $2$ ordered pairs satisfying this equation.

if $m$ is odd number then $(m-8)(m-10)$ is odd number doesn't satisfy $2^{n}$ so $m$ is even number.

let $m = 2k$ where $k$ is natural number than: $(m-8)(m-10) = (2k-8)(2k-10) = 4(k-4)(k-5) = 2^{n}$ for satisfying this we need: $(k-4)$ and $(k-5)$ having the same sign and

$( |(k-4)| = 1$ and $|(k-5)| = 2^{n-2})$

or

$(|(k-4)| = 2^{n-2}$ and $|(k-5)| = 1)$

so $k=3$ and $k=6$ satisfy it $\Rightarrow m = 6$ and $m=12$

we have two factors (m-8) and (m-10).to satisfy the given condition,m should be such that both the factors give numbers which are (2,4,8 ,16.......).only 6 and 12 satisfy the condition.thus only two order pairs can be formed