Find all positive integers s.t. $10^m-8^n=2m^2$ Find all pairs of positive integers $(m,n)$ such that $10^m-8^n=2m^2$

Hint:
$${10}^m - 8^n = {2^m}{5^m} - {2^{3n}}$$
so that
$$m^2 = {2^{m-1}}{5^m} - {2^{3n-1}}.$$
In particular:
$${2^{m-1}}{5^m} - {2^{3n-1}} \geq 1,$$
and
$${2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$
is a square, which means $m \equiv 1 \pmod 2$.
Suppose that $m > 1$.  Then the RHS of
$$m^2 = {2^{m-1}}\left({5^m} - {2^{3n-m}}\right)$$
is even while the LHS is odd (since $5^m - {2^{3n-m}}$ is always odd, unless $m = 3n$ [I'll deal with this case later]).  A contradiction.  Therefore, $m = 1$.
Substituting $m = 1$:
$$10 - {8^n} = 2$$
which gives $n = 1$.
Update:  If $m = 3n$, then
$$m^2 = {3^2}{n^2} = 2^{3n - 1}\left(5^{3n} - 1\right).$$
Since $\gcd(3,2) = 1$, then it follows that
$$2^{3n - 1} \mid n^2$$
which implies that $2^{3n - 1} \leq n^2$, contradicting $n^2 < 2^{3n - 1}$.
Therefore, in general, $m = 3n$ cannot occur.
Hence the only solution to the original problem is $(m, n) = (1, 1)$.
