We know that $|12^1 - 5^1| = 7$. I will prove that this is the minimum value. Claim: There are no solutions to $|12^m - 5^n| = k$ if $k < 7$. Proof. Suppose there exists a nonnegative integer $k < 7$ that satisfies $|12^m - 5^n| = k$. Since $12^m$ is even and $5^n$ is odd, the distance between them must be odd. Thus, $k$ must be $1$, $3$, or $5$. Clearly, $3$ and $5$ don't work because $3 \nmid 5^n$ and $5 \nmid 12^m$. Thus, $k=1$. We must have $12^m - 5^n = \pm 1 \implies 5^n = 12^m \pm 1$. Case 1: $5^n = 12^m - 1$. By the difference of nth powers factorization, the RHS is divisible by $11$, but the LHS is not, contradiction. Case 2: $5^n = 12^m + 1$. Idk how to do this.

Case 2: $5^n = 12^m + 1$. This means $2^m + 1 \equiv 0 \pmod{5}$, which means $m \equiv 2 \pmod{4}$. We can let $m = 4p - 2$ for some positive integer $p$. Thus, we have: $$12^{4p - 2} + 1 = 5^n$$$$144^{2p - 1} + 1 = 5^n$$By sum of nth powers factorization, the LHS is divisible by $145$, meaning it is divisible by $29$, contradiction.

For case 2, we have $$5^n-1=12^m$$Note that $5^n-1$ is only divisibly by $12$ when $n$ is even (take $\mod 3$ and $\mod 4$) Moreover by LTE, $v_2(5^{2k}-1)=3+v_2(k)$ and $v_3(5^{2k}-1)=1+v_3(k)$ For this to be a power of 12, a requirement is that $v_2(5^{2k}-1)=2\cdot v_3(5^{2k}-1)$, $$3+v_2(k)=2+2v_3(k)$$$$1+v_2(k)=2 v_3(k)$$That means that $k$ can be expressed in the form $2^{2a-1}3^a\ldots$. This implies that $12|n$. However, if $12|n$, we have that $5^n-1\equiv 0\mod 13$, hence it is not a power of $12$.