Example 3 Find all integer solutions of the indeterminate equation
$$
\left(x^{2}-4\right)\left(y^{2}-4\right)=t^{2}+12
$$
When $|x| \leqslant 2$, we can obtain
$$
(x, y, t)=(0,0, \pm 2),(0, \pm 1,0),( \pm 1,0,0) \text {. }
$$

The following proof shows that these are the only solutions.
When $|x| \geqslant 3$, without loss of generality, assume $x, y, t \geqslant 0$.
It is easy to see that $y \geqslant 3, 3 \nmid x, 3 \nmid y$.
Then $x^{2} \equiv y^{2} \equiv 1(\bmod 3)$
$$
\begin{array}{l}
\Rightarrow 9\left|\left(x^{2}-4\right)\left(y^{2}-4\right) \Rightarrow 9\right|\left(t^{2}+12\right) \\
\Rightarrow 9\left|\left(t^{2}+3\right) \Rightarrow 3\right| t \Rightarrow 9 \mid 3,
\end{array}
$$

which is a contradiction.
Therefore, at least one of $3 \mid x$ or $3 \mid y$ must hold.
Without loss of generality, assume $3 \mid x$.
Thus, $x^{2}-4$ is a $3k-1$ type number, and it has at least one $3k-1$ type prime factor $p$.

If $p$ is an odd prime, then $p$ is a $6k-1$ type prime, and we have $p \mid\left(t^{2}+12\right)$, which contradicts Corollary 2.
Thus, $p$ must be the even prime 2.
Therefore, $2 \mid \left(x^{2}-4\right)$, and $x$ is even.
Let $x=2 x^{\prime}\left(x^{\prime} \in \mathbf{Z}\right)$.
Thus, $4\left(x^{\prime 2}-1\right)\left(y^{2}-4\right)=t^{2}+12$.
Similarly, $2 \mid \left(x^{\prime 2}-1\right)$.
Therefore, $x^{\prime}$ is odd, and $t^{2}+12 \equiv 0(\bmod 32)$.
This is impossible, leading to a contradiction.
In conclusion, the solutions obtained earlier are all the solutions to the equation.