Example 18 (25th All-Soviet Union Mathematical Olympiad) Find the integer solutions of the system of equations $\left\{\begin{array}{l}x z-2 y t=3, \\ x t+y z=1 .\end{array}\right.$
From the system of equations, we can obtain $(x z-2 y t)^{2}+2(x t+y z)^{2}=\left(x^{2}+2 y^{2}\right)\left(z^{2}+2 t^{2}\right)=11$,
thus we have $x^{2}+2 y^{2}=1$ or $z^{2}+2 t^{2}=1$.
If $x^{2}+2 y^{2}=1$, then we can get $y=0, x= \pm 1$, and from the second equation, we get $t= \pm 1$, and from the first equation, we get $z= \pm 3$.
If $z^{2}+2 t^{2}=1$, then we can get $t=0, z= \pm 1$, and then $y= \pm 1, x= \pm 3$.
Direct verification shows that the four sets of solutions $(x, y, z, t)$ satisfy the system of equations, they are $(1,0,3,1)$,
$$
(-1,0,-3,-1),(3,1,1,0),(-3,-1,-1,0)
$$