For $m=15$ and $n=2$, the expression $f(m,n):=|2^m - 181^n|$ takes the value $7$. As $2^m$ is even and $181^n$ is odd, we get that $f(m,n)$ is odd; this eliminates the candidate values $0,2,4,6$. It remains to eliminate the three values $1,3,5$. Since $f(1,n)=181^n-2\ge179$, we may assume $m\ge2$ so that $2^m$ is a multiple of $4$. (1) Suppose $f(m,n)=1$. Arguing modulo $4$, this implies $181^n-2^m=1$, or equivalently $181^n-1=2^m$. Since $181^n-1$ is a multiple of $5$, also $2^m$ should be a multiple of $5$; contradiction. (2) Suppose $f(m,n)=3$. Arguing modulo $4$, this implies $2^m-181^n=3$. ??? This case is still open. ??? (3) Suppose $f(m,n)=5$. Arguing modulo $4$, this implies $181^n-2^m=5$. Arguing modulo $5$, this implies $m=4k$. But then $181^n-2^{4k}$ is a multiple of $3$, whereas $5$ is not; contradiction.

The case $f(m, n) = 3$ is settled by considering $f(m, n)$ mod $7, 9$.

$m=15$ and $n=2$ gives the value $7$. So we show that $0, 1, 2 ,3 ,4, 5, 6$ are impossible. $0, 2, 4, 6$: Impossible by $\pmod{2}$. $1$. Either $2^m - 181^n = 1$ or $181^n - 2^m = 1$. The first is impossible by mod 4, the second by mod 5. $3$. Either $2^m-181^n=3$ or $181^n-2^m=3$. The second is impossible by mod 4. $5$. Either $2^m-181^n=5$, or $181^n-2^m=5$. The first is impossible by mod 4. For $2^m-181^n=3$, by mod $9$ we have $2^m \equiv 4 \pmod {9}$, so $m \equiv 2 \pmod {6}$. This shows $4 \cdot 64^k - 181^n = 3$, so by mod $7$ we have $n$ even, so $2^m - 181^n$ is a difference of two squares. For $181^n - 2^m = 5$, by mod 5 we have $m \equiv 0 \pmod{4}$, and then mod 3 does the rest.

$m=15$ and $n=2$ gives the value $7$. $|2^m - 181^n|$ is odd number. So we show that $0, 1, 2 ,3 ,4, 5, 6$ are impossible. 1.$|2^m - 181^n|=\mp 1$ zsigmondy theorem.contradiction. 2.a)$2^m - 181^n=3$ by mod3 $m$ is evan.$n$ is odd. $$ 181^n \equiv (-1)^n \equiv -1 mod(13) $$$2^m=x^2$ $$x^2= 181^n+3 \equiv 2 mod(13) $$$$(\frac{2}{13})=1=(-1)^{\frac{13^2-1}{8}}=-1$$the rest are simple mod 4 and mod5

We need to check $1, 3, 5$. Due to Mihailescu theorem we can exclude $1$ case. Assume $ | 2^m - 181^n | = 0 \mod 5$. Since $181 = 1 \mod 5$, so $m=4k$. However $181 = 1 \mod 3$, so $ | 2^m - 181^n | = 0 \mod 3$, thus $| 2^m - 181^n | \neq 5$. Thus if $ | 2^m - 181^n | = 0 \mod 3$, then $m=2(2l+1)$. It implies that $| 2^m - 181^n | = 1, 4 \mod 5$, which also implies $| 2^m - 181^n | \neq 3 $. Hence for $m=15, n=2$ we have the minimum value $7$.