Example 18 (25th All-Soviet Union Mathematical Olympiad) Find the integer solutions of the system of equations:
$$
\left\{\begin{array}{l}
x z-2 y t=3, \\
x t+y z=1
\end{array}\right.
$$
Solving the original system of equations by squaring both sides, we get
$$
\left\{\begin{array}{l}
x^{2} z^{2}-4 x y z t+4 y^{2} t^{2}=9 \\
x^{2} t^{2}-2 x y z t+y^{2} z^{2}=1
\end{array}\right.
$$
(1) + 2 × (2), simplifying yields $\left(x^{2}+2 y^{2}\right)\left(z^{2}+2 t^{2}\right)=11$.

Thus, $x^{2}+2 y^{2}=1$ or $z^{2}+2 t^{2}=1$.
(1) From $x^{2}+2 y^{2}=1$, we get $y=0, x= \pm 1$, and from the original system of equations, we get $z= \pm 3, t= \pm 1$.
(2) From $z^{2}+2 t^{2}=1$, we get $t=0, z= \pm 1$, and from the original system of equations, we get $x= \pm 3, y= \pm 1$.
Upon verification, the original system of equations has only 4 solutions, which are $(x, y, z, t)=(1,0,3,1),(-1,0,-3$,
$$
-1),(3,1,1,0),(-3,-1,-1,0) \text {. }
$$