4. If $m$ and $n$ are both positive integers, find the minimum value of $\left|2^{m}-181^{n}\right|$.

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4. When $m=15, n=2$, we have $\left|2^{m}-181^{n}\right|=\left|2^{15}-181^{2}\right|=7$.
Assume there exist positive integers $m, n$ such that $\left|2^{m}-181^{n}\right|<7$.
Then, by $12^{m}-181^{n} \mid$ being odd, we know $\left|2^{m}-181^{n}\right|=1$ or 3 or 5.
Note that, $2^{m}$ modulo 15 can only be $1, 2, 4, 8$, and $181^{n} \equiv 1(\bmod 15)$.
Thus, $2^{m}-181^{n}$ modulo 15 can only be $0, 1, 3, 7$.
Therefore, $2^{m}-181^{n}=1$ or 3.
If $2^{m}-181^{n}=1$, then $2^{m} \equiv 2(\bmod 4)$. Hence, $m=1$, which has no solution.
If $2^{m}-181^{n}=3$, then
$2^{m} \equiv 1(\bmod 3) \Rightarrow m$ is even;
$2^{m} \equiv 4(\bmod 5) \Rightarrow m \equiv 2(\bmod 3)$.
Thus, $m \equiv 2(\bmod 6)$.
Let $m=6 k+2(k \in \mathbf{N})$. Then
$2^{6 k+2}=181^{n}+3$.
$\Rightarrow 4 \times(-1)^{k} \equiv(-1)^{n}+3(\bmod 13)$.
Hence, $k$ and $n$ are both even. Let $n=2 n_{1}\left(n_{1} \in \mathbf{Z}_{+}\right)$.
Then $3=\left|2^{6 k+2}-181^{n}\right|=\left|2^{6 k+2}-181^{2 n_{1}}\right|$
$=\left|\left(2^{3 k+1}+181^{n_{1}}\right)\left(2^{3 k+1}-181^{n_{1}}\right)\right|$
$\geqslant 2^{3 k+1}+181^{n_{1}} \geqslant 183$,

a contradiction, no solution.
In summary, the minimum value of $\left|2^{m}-181^{n}\right|$ is 7.