Example 3 Given positive integers $k, l$. Find all functions $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, such that for any $m, n \in \mathbf{Z}_{+}$, we have $(f(m)+f(n)) \mid (m+n+l)^{k}$.
【Analysis】First, when $l$ is odd, take $m=n$, we have
$$
2 f(m) \mid(2 m+l)^{k},
$$

which is obviously a contradiction.
Therefore, when $l$ is odd, there does not exist a function $f$ that satisfies the condition.
Next, we discuss the case when $l$ is even.
We still conjecture that $f(n)=n+\frac{l}{2}$.
To do this, we first prove a lemma.
Lemma For any $m \in \mathbf{Z}_{+}$, we have
$$
f(m+1)-f(m)= \pm 1 \text {. }
$$

Proof Clearly, $f$ is not a constant function.
If $|f(m+1)-f(m)| \neq 1$, then there exists a prime $p$,
such that
$$
p \mid(f(m+1)-f(m)) .
$$

Take $e$ such that $p^{\circ}>m+l$. Combining the condition, we have
$$
\begin{array}{l}
\left(f\left(p^{e}-m-l\right)+f(m)\right) \mid\left(p^{e}-m-l+m+l\right)^{k} \\
\Rightarrow\left(f\left(p^{e}-m-l\right)+f(m)\right) \mid p^{e k} . \\
\text { Also, } f\left(p^{e}-m-l\right)+f(m) \geqslant 1+1=2, \text { so } \\
p \mid\left(f\left(p^{e}-m-1\right)+f(m)\right) .
\end{array}
$$

By the assumption, we get
$$
\begin{array}{l}
p \mid\left(f\left(p^{e}-m-1\right)+f(m+1)\right) . \\
\text { And }\left(f\left(p^{e}-m-1\right)+f(m+1)\right) \mid\left(p^{e}+1\right)^{k},
\end{array}
$$

which is clearly a contradiction.
The lemma is proved.
By the lemma, we have
$$
f(n) \leqslant f(n-1)+1 \leqslant \cdots \leqslant f(1)+(n-1) \text {. }
$$

Next, choose a sufficiently large prime $p$, satisfying
$$
p>l+1, p^{2}>p+2 f(1)-l-2 \text {. }
$$

Notice,
$$
\begin{array}{l}
(f(1)+f(p-l-1)) \mid(p-l-1+l+1)^{k} \\
\Rightarrow\left(f(1)+f(p-l-1) \mid p^{k},\right. \\
q)$. Then $f(p-l-1)=p-f(1)$,
and $f(q-l-1)=q-f(1)$.
Thus, $p-f(1)=f(p-l-1)$
$\leqslant f(p-l-2)+1 \leqslant \cdots$
$\leqslant f(q-l-1)+(p-q)$
$=q-f(1)+(p-q)=p-f(1)$.
All equalities in the above expression hold, indicating that for $q \leqslant n \leqslant p$, we have
$f(n-l-1)=n-f(1)$.
In particular, for $n \geqslant q-l-1$, we have
$f(n)=n+l+1-f(1)=n+c$ (where $c$ is a constant).
Fix $n$, and choose
$$
q-l-1 \leqslant N \leqslant p-l-1,
$$

such that $n+N+l$ is a prime.
In fact, since $p$ is sufficiently large, we have
$(f(n)+f(N)) \mid(n+N+l)^{k}$
$\Rightarrow f(n)+f(N)=(n+N+l)^{e}(1 \leqslant e \leqslant k)$.
Also, $f(N)=N+c$, and as $N \rightarrow+\infty$, unless $e=1$, a contradiction will inevitably arise. Therefore,
$f(n)=n+l-c\left(n \in \mathbf{Z}_{+}\right)$.
Substituting into the condition of the problem, we get
$$
(m+n+2 l-2 c) \mid(m+n+l)^{k} \text {. }
$$

Then clearly,
$$
\begin{array}{l}
(m+n+2 l-2 c) \mid(m+n+l-(m+n+2 l-2 c))^{k} \\
\Rightarrow(m+n+2 l-2 c) \mid(2 c-l)^{k} .
\end{array}
$$

Taking $m \rightarrow+\infty, n \rightarrow+\infty$, the right side is a constant, which must be 0, i.e., $c=\frac{l}{2}$.
Therefore, when $l$ is even, $f(n)=n+\frac{l}{2}\left(n \in \mathbf{Z}_{+}\right)$; when $l$ is odd, there is no function $f$ that satisfies the condition.