Example 3 Given a sequence of positive integers $\left\{a_{n}\right\}$ satisfying:
$$
a_{0}=m, a_{n+1}=a_{n}^{5}+487(n \in \mathbf{N}) \text {. }
$$

Find the value of the positive integer $m$ such that the number of perfect squares in the sequence $\left\{a_{n}\right\}$ is maximized.
Note that if $a_{n}$ is a perfect square, then $a_{n} \equiv 0,1(\bmod 4)$.
If $a_{0} \equiv 0(\bmod 4)$, then
$$
\begin{array}{l}
a_{1} \equiv 3(\bmod 4), a_{2} \equiv 2(\bmod 4), \\
a_{3} \equiv 3(\bmod 4), \cdots ;
\end{array}
$$

If $a_{0} \equiv 1(\bmod 4)$, then
$$
\begin{array}{l}
a_{1} \equiv 0(\bmod 4), a_{2} \equiv 3(\bmod 4), \\
a_{3} \equiv 2(\bmod 4), \cdots ;
\end{array}
$$

If $a_{0} \equiv 2(\bmod 4)$, then
$$
\begin{array}{l}
a_{1} \equiv 3(\bmod 4), a_{2} \equiv 2(\bmod 4), \\
a_{3} \equiv 3(\bmod 4), \cdots ;
\end{array}
$$

If $a_{0} \equiv 3(\bmod 4)$, then
$$
\begin{array}{l}
a_{1} \equiv 2(\bmod 4), a_{2} \equiv 3(\bmod 4), \\
a_{3} \equiv 2(\bmod 4), \cdots .
\end{array}
$$

Therefore, the number of perfect squares in the sequence $\left\{a_{n}\right\}$ is more than 2.

If the number of perfect squares in the sequence $\left\{a_{n}\right\}$ is 2, then $a_{0} \equiv 1(\bmod 4)$.
Let $a_{0}=4 k+1=s^{2}$. Then $a_{1}=s^{10}+487=t^{2}$.
Thus, $487=\left(t+s^{5}\right)\left(t-s^{5}\right)$.
Since 487 is a prime number, we have,
$$
t-s^{5}=1, t+s^{5}=487 \text {. }
$$

Solving these, we get $t=244, s=3$.
Thus, $m=a_{0}=9, a_{1}=244^{2}$.
Therefore, the required positive integer $m=9$.