Example 3. If $n$ is a natural number, and $9 n^{2}+5 n+26$ equals the product of two consecutive natural numbers, find $n$. (1985 Shanghai Junior High School Mathematics Competition Question)
Let $M=9 n^{2}+5 n+26$, then
$$
\begin{aligned}
4 M+1 & =4\left(9 n^{2}+5 n+26\right)+1 \\
& =36 n^{2}+20 n+105 .
\end{aligned}
$$

Let $4 M+1=(6 n+k)^{2}$, where $k$ is a natural number, we get
$$
36 n^{2}+20 n+105=36 n^{2}+12 k n+k^{2},
$$

Solving for $n$ gives $n=\frac{k^{2}-105}{20-12 k}$.
From $n>0$ we get $\frac{5}{3}<k<\sqrt{105}$, i.e., $2 \leqslant k \leqslant 10$.
Substituting values of $k$ into equation (1) one by one, we get two solutions that meet the conditions:
When $k=3$, $n=6$,
at this time $9 n^{2}+5 n+26=380=19 \times 20$,
When $k=5$, $n=2$,
at this time $9 n^{2}+5 n+26=72=8 \times 9$.