Example 1 Find all functions $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ such that for all $m, n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
f(m n)=f(m) f(n), \\
(m+n) \mid(f(m)+f(n)) .{ }^{[1]}
\end{array}
$$
(2016, Turkey National Team Selection Exam)
【Analysis】First, try to guess the answer to the problem.
Substitute $m=n=1$ into equation (1), we have $f(1)=f^{2}(1)$.
Combining $f(n) \in \mathbf{Z}_{+}$, we get $f(1)=1$.
Substitute $n=1$ into equation (2), we have
$$
(m+1) \mid(f(m)+1) \text {. }
$$

Thus, we conjecture: $f(m)=m^{k}$ ( $k$ is an odd number).
Now, we will prove that the conjecture is correct.
$$
\begin{array}{l}
\text { By }(2 n+1) \mid(f(2 n)+f(1)) \\
\Rightarrow(2 n+1) \mid(f(2) f(n)+1) \\
\Rightarrow(2 n+1, f(2) f(n))=1 \\
\Rightarrow(2 n+1, f(2))=1\left(n \in \mathbf{Z}_{+}\right) \\
\Rightarrow f(2)=2^{k} .
\end{array}
$$

By $(1+2) \mid(f(1)+f(2))$
$\Rightarrow 3 \mid(f(2)+1) \Rightarrow k$ is an odd number.
From equation (1), we know that for any $m \in \mathbf{Z}_{+}$, we have
$$
f\left(2^{m}\right)=f^{m}(2)=2^{k m} \text {. }
$$

Combining that $k$ is an odd number, we have
$$
\left(2^{m}+n\right) \mid\left(2^{m k}+n^{k}\right)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

From equation (2), we know
$$
\begin{array}{l}
\left(2^{m}+n\right) \mid\left(f\left(2^{m}\right)+f(n)\right) \\
\Rightarrow\left(2^{m}+n\right) \mid\left(2^{k m}+f(n)\right) \\
\Rightarrow\left(2^{m}+n\right) \mid\left(f(n)-n^{k}\right) .
\end{array}
$$

By the arbitrariness of $m$, we have
$$
f(n)-n^{k}=0 \Rightarrow f(n)=n^{k}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Conversely, it is easy to verify that it satisfies the requirements of the problem.
In summary, $f(n)=n^{k}\left(n \in \mathbf{Z}_{+}, k\right.$ is a positive odd number $)$, which is the required $f$.