Example 17 (Problem 1471 from "Mathematics Bulletin") Find the non-negative integer solutions of the system of equations $\left\{\begin{array}{l}x+y=z t \\ z+t=x y\end{array}\right.$.
Solve: Since in the system of equations, $x$ and $y$, $z$ and $t$ can be interchanged, we can first find the integer solutions $(x, y, z, t)$ that satisfy $0 \leqslant x \leqslant y, 0 \leqslant z \leqslant t$.
(1) If one of $x$ and $z$ is zero, let's assume $x=0$, then by eliminating $t$ from the original system of equations, we get: $y+z^{2}=0$ so $y=z=0, t=0$. Thus, $(0,0,0,0)$ is one solution to the original system of equations.
(2) If $x$ and $z$ are both not 0, but one of them is 1, let's assume $x=1$, then by eliminating $y$ from the original system of equations, we get: $t+z=zt-1$ so $(z-1)(t-1)=2$. Since $z, t$ are positive integers and $z \leqslant t$, we have $\left\{\begin{array}{l}z-1=1 \\ t-1=2\end{array}\right.$, which gives $z=2, t=3, y=5$. Thus, $(1,5,2,3)$ is one solution to the original system of equations. Similarly, $(2,3,1,5)$ is also a solution to the original system of equations.
(3) If $x \geqslant 2$ and $z \geqslant 2$, from $(x-1)(y-1) \geqslant 1$ we get $xy \geqslant x+y$.
Equality holds if and only if $x=y=2$. Similarly, $zt \geqslant z+t$.
Equality holds if and only if $z=t=2$.
From (1), (2), and the original system of equations, we get: $xy \geqslant z+t$,
Equality holds if and only if $x=y=z=t=2$.
Therefore, the original system of equations has only one solution $(2,2,2,2)$ in this case.
According to the symmetry of the system of equations, we can also get that $(5,1,2,3), (5,1,3,2), (1,5,3,2), (2,3,5,1), (3,2,1,5), (2,3,1,5)$ are also solutions to the original system of equations.

In summary, the original system of equations has 10 non-negative integer solutions: $(0,0,0,0), (2,2,2,2), (1,5,2,3), (5,1,2,3), (1,5,3,2), (5,1,3,2), (2,3,1,5), (2,3,5,1), (3,2,1,5), (3,2,5,1)$.