Example 4 Given that $x$ and $y$ are positive integers, and satisfy the conditions $x y + x + y = 71, x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$.
Solution: Let $x+y=u, xy=v$. From the given equation, we get $u+v=71, uv=880$.
By Vieta's formulas, $u$ and $v$ are the roots of the quadratic equation
$$
t^{2}-71t+880=0
$$

Solving this equation, we get $t=16$ or $t=55$.
Therefore, $\left\{\begin{array}{l}u=16, \\ v=55\end{array}\right.$ or $\left\{\begin{array}{l}u=55, \\ v=16,\end{array}\right.$ which means $\left\{\begin{array}{l}x+y=16, \\ xy=55\end{array}\right.$ or $\left\{\begin{array}{l}x+y=55, \\ xy=16 .\end{array}\right.$

From the first set of equations, we get $x^{2}-16x+55=0$, $\Delta=6^{2}$, the equation has integer roots;

From the second set of equations, we get $x^{2}-55x+16=0$, $\Delta=2961$, the equation has no integer roots.
Only $x+y=16, xy=55$ meets the condition.
Thus, $x^{2}+y^{2}=(x+y)^{2}-2xy=146$.