Easy :Suppose there exist $ k < 7$ and the equation has solution . $ |12^m - 5^n|\equiv 1(\mod 2)$ and $ 3,5\not|12^m - 5^n$ so we must prove for $ k = 1$ Suppose $ |12^m - 5^n| = 1$ Case 1$ 12^m - 5^n = 1$ But $ 5^m + 1\equiv 2(\mod 4)$ then the equation has no solution. Case 2$ 12^m - 5^n = - 1$ $ 5^n - 1\equiv ( - 1)^n - 1 (\mod 2)$ so $ n = 2p$ $ 12^m = (5^p - 1)(5^p + 1)$ has no solution! This problem was be prove.

TTsphn wrote: $ 12^m = (5^p - 1)(5^p + 1)$ has no solution! This is quite unclear and seems to skip many lines of proof.

Let $N = \left| {{{12}^m} - {5^n}} \right|$, since $(N,30) = 1$, so if $N \leqslant 6$, then $N = 1$. Because ${5^n} \equiv 5,3,4,9,1(\bmod 11)$, so ${12^m} - {5^n}\not \equiv \pm 1(\bmod 11)$, hence $N \ne 1$, contradiction. So $N \geqslant 7$.

zsigmondy theorem

Of course, when $m = 1$, we get the solution $12^1 - 5^1 = 7$. I claim that $| 12^m - 5^n| \ge 19$ if $m \ge 2$. To see this, we are going to view the equation modulo $3120$; for $m \ge 2$ we have $12^m \pmod{3120} \in \{144, 1728, 2016, 2352\} = A$ while $5^n \pmod{3120} \in \{5, 25, 125, 625\} = B$. Since the integers in $A$ and $B$ that are closest to one another are $125$ and $144$ we get $|12^m - 5^n| \ge 19$ for all $m, n \in \mathbb{N}$ with $m \ge 2$.