Example 6 Find all prime numbers $p$ such that $4 p^{2}+1$ and $6 p^{2}+1$ are also prime.
Let $u=4 p^{2}+1, v=6 p^{2}+1$,
and let $p=5 k+r, k \in \mathbf{Z}, r \in\{0,1,2,3,4\}$.
Then
$$
\begin{array}{l}
u=100 k^{2}+40 k r+4 r^{2}+1, \\
v=150 k^{2}+60 k r+6 r^{2}+1 .
\end{array}
$$

If $p \equiv 0(\bmod 5)$, then $u \equiv 1, v \equiv 1(\bmod 5)$;
if $p \equiv 1(\bmod 5)$, then $u \equiv 0, v \equiv 2(\bmod 5)$;
if $p \equiv 2(\bmod 5)$, then $u \equiv 2, v \equiv 0(\bmod 5)$;
if $p \equiv 3(\bmod 5)$, then $u \equiv 2, v \equiv 0(\bmod 5)$;
if $p \equiv 4(\bmod 5)$, then $u \equiv 0, v \equiv 2(\bmod 5)$.
Therefore, for any integer $p$, among the three numbers $p, u, v$, there is one and only one that can be divisible by 5.
Since when $p$ is not divisible by 5, one of $u$ and $v$ is divisible by 5, i.e., one of $u$ and $v$ is not a prime number, so for $u$ and $v$ to both be prime numbers, it is only possible when $p$ is divisible by 5. Since $p$ is a prime number, $p=5$.
It is not difficult to verify that when $p=5$, $u=4 \cdot 5^{2}+1=101, v=6 \cdot 5^{2}+1=151$ are both prime numbers. Therefore, this problem has only one solution.