$$\equiv m^{n}+n^{m} \equiv 0\left(\bmod p_{i}\right) .$$

On the other hand,
$$\begin{array}{l} m^{n_{1}}+n_{1}^{m} \\ =m^{n}+m^{n}\left[\left(m^{\varphi\left(p_{j}^{p}\right)}\right)^{\varphi}\left(\left(\frac{p_{1} p_{2}-p_{k}}{p_{i}}\right)^{c}\right)-1\right]+ \\ \quad n^{m}+\sum_{j=1}^{m} \mathrm{C}_{m}^{j} n^{m-j} M^{j}, \end{array}$$
$$v_{p_{i}}\left(\sum_{j=1}^{m} \mathrm{C}_{m}^{j} n^{m-j} M^{j}\right) \geqslant v_{p_{i}}(M) \geqslant t-1 \text {, }$$

Therefore, $v_{p_{i}}\left(m^{n}\left[\left(m^{\varphi\left(p_{i}^{t}\right)}\right)^{\varphi\left(\left(\frac{p_{1} p_{2} \cdots p_{k}}{p_{i}}\right)^{t}\right)}-1\right]+\right.$
$$\left.\sum_{j=1}^{m} \mathrm{C}_{m}^{j} n^{m-j} M^{j}\right) \geqslant t-1 \text {. }$$

But $v_{p_{i}}\left(m^{n}+n^{m}\right) \leqslant \max _{1 \leqslant i \leqslant k}\left\{\boldsymbol{\alpha}_{i}\right\}m^{n}+n^{m}$, so $m^{n_{1}}+n_{1}^{m}$ has a prime factor different from $p_{1}, p_{2}, \cdots, p_{k}$, meaning it has at least $k+1$ distinct prime factors.
Thus, the conclusion holds for $k+1$.
In summary, for any positive integer $k$, there exists a positive integer $n$ such that $m^{n}+n^{m}$ has at least $k$ distinct prime factors.
Proof by induction on $k$.
When $k=1$, take $n=m$.
Assume the conclusion holds for $k$, i.e., there exists a positive integer $n$ such that $m^{n}+n^{m}$ has at least $k$ distinct prime factors.
Now prove the case for $k+1$.
If $m^{n}+n^{m}$ has at least $k+1$ distinct prime factors, then the conclusion holds for $k+1$.
If $m^{n}+n^{m}$ has exactly $k$ distinct prime factors, let $m^{n}+n^{m}=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$.
Take $t>\max _{1 \leqslant i \leqslant k}\left\{\alpha_{i}\right\}+1 \geqslant 2 \in \mathbf{Z}_{+}$,
$n_{1}=n+\varphi\left(\left(p_{1} p_{2} \cdots p_{k}\right)^{t}\right)$.
Let $M=\varphi\left(\left(p_{1} p_{2} \cdots p_{k}\right)^{t}\right)$.
By Euler's theorem,
\begin{array}{l} m^{n_{1}}+n_{1}^{m}=m^{n+M}+(n+M)^{m} \\ \equiv m^{n}\left(m^{\varphi\left(p_{i}^{t}\right)}\right)^{\varphi}\left(\left(\frac{p_{1} p_{2} \cdots p_{k}}{p_{i}}\right)^{t}\right)+ \\ \quad\left[n+\left(p_{1} p_{2} \cdots p_{k}\right)^{t-1} \prod_{i=1}^{k}\left(p_{i}-1\right)\right]^{m}