Example 4 Find all positive integers $k$ such that there exist positive integers $m, n$ satisfying
$$
m(m+k)=n(n+1) \text {. }
$$
Solution: Clearly, $k=1$ meets the requirement, at this time, $m=n$.
When $k>1$, it is obvious that $m>3$.
When $k \geqslant 4$, from equation (1) we get
$$
\begin{array}{l}
\left(4 m^{2}+4 m k+k^{2}\right)-k^{2}+1=4 n^{2}+4 n+1 \\
\Rightarrow(2 m+k)^{2}-(2 n+1)^{2}=k^{2}-1 .
\end{array}
$$

By the difference of squares formula, we get
$$
\begin{array}{l}
(2 m+k+2 n+1)(2 m+k-2 n-1) \\
=k^{2}-1 . \\
\text { Let }\left\{\begin{array}{l}
2 m+k+2 n+1=k^{2}-1, \\
2 m+k-2 n-1=1 .
\end{array}\right.
\end{array}
$$

Solving, we get $(m, n)=\left(\frac{k^{2}-2 k}{4}, \frac{k^{2}}{4}-1\right)$.
To make $m, n$ both integers, $k$ should be even, let $k=2 t\left(t \in \mathbf{Z}_{+}, t>1\right)$.
Then $(m, n)=\left(t^{2}-t, t^{2}-1\right)$.
If $k$ is odd, let $k=2 t+1\left(t \in \mathbf{Z}_{+}, t>1\right)$. Then equation (2) becomes
$$
\begin{array}{l}
(2 m+2 t+2 n+2)(2 m+2 t-2 n) \\
=4 t^{2}+4 t . \\
\text { Let }\left\{\begin{array}{l}
2 m+2 t+2 n+2=2 t^{2}+2 t, \\
2 m+2 t-2 n=2 .
\end{array}\right.
\end{array}
$$

Solving, we get $(m, n)=\left(\frac{t^{2}-t}{2}, \frac{t^{2}+t}{2}+1\right)$.
It is easy to see that when $t$ is an integer, $m, n$ are both integers.
In summary, when $k=1$ or $k \geqslant 4$, there exist positive integers $m, n$ that satisfy equation (1).