Let $N=\mid 12^m - 5^n\mid$. It is clear that $N$ is an odd number and, neither a multiple of $3$ nor a multiple of $5$. This means that $N=1$ or $N\ge 7$.

First, let's show $N\ne 1$.

• If $(12^m - 5^n) = 1$, then taking MOD $4$ on both sides yields $-1\equiv 1\pmod{4}$. It cannot hold.
• If $(5^n - 12^m) = 1$, then taking MOD ${3}$ on both sides yields $(-1)^n\equiv 1\pmod{3}$. Therefore $n$ must be even. Taking MOD $5$ on the original expression gives $2^m \equiv -1\pmod{5}$ which means $2^{m+2}\equiv 1\pmod{5}$. Letting $n=2k$ and $m+2=4l$ will lead to the following relation which is contradicting. $$5^n - 12^m = 5^{2k} - 12^{4l}=\left(5^k + 12^{2l}\right)\left(5^k-12^{2l}\right) > 1$$

Therefore, we conclude $N$ cannot equal $1$. Hence, $N\ge 7$. Clearly, when $m=n=1$, we have $N = 7$. This means that the answer is $\boxed{7}$.