Example 5 Find the integer solutions of the equation $x^{2}-15 y^{2}=61$.

保持源文本的换行和格式。
First, find the integer solution suitable for
$$
l^{2} \equiv 15(\bmod 61)\left(0 \leqslant l \leqslant \frac{61}{2}\right)
$$

which means solving the equation
$$
l^{2}=15+61 h\left(l^{2} \leqslant 900\right)
$$

Since $0 \leqslant h \leqslant\left[\frac{900}{61}\right]=14$, we test $h=0,1, \cdots, 14$ one by one, and find that when $h=10$,
$$
15+61 \times 10=625=25^{2} \text {. }
$$

At this point, $l=25$.
Next, we find the integer solution to the equation $x_{1}^{2}-15 y_{1}^{2}=10$, which is to find the integer solution to the equation
$$
l^{2}=15+10 h\left(l \leqslant \frac{10}{2}=5\right)
$$

By testing, we find that $l=5, h=1$.
Thus, $x_{1}=5, y_{1}=1$.
Then, we find the solution to the equation $x_{2}^{2}-15 y_{2}^{2}=1$.
This is a Pell's equation, and its smallest solution is
$$
x_{2}=4, y_{2}=1 \text {. }
$$

Therefore, $x_{2}+\sqrt{15} y_{2}= \pm(4+\sqrt{15})^{n}$,
$$
\begin{array}{l}
x_{1}+\sqrt{15} y_{1}= \pm(4+\sqrt{15})^{n}(5 \pm \sqrt{15}), \\
x+\sqrt{15} y \\
= \pm(4+\sqrt{15})^{n}(5 \pm \sqrt{15})\left(\frac{25 \pm \sqrt{15}}{10}\right) .
\end{array}
$$

Since the three $\pm$ signs are independent, the solutions to the equation are
$$
x+\sqrt{15} y= \pm(4+\sqrt{15})^{n}(14 \pm 3 \sqrt{15})
$$

and $x+\sqrt{15} y= \pm(4+\sqrt{15})^{n}(11 \pm 2 \sqrt{15})$.