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 $x^{2} + y^{2}$.
(Jiangsu Province 14th Junior High School Mathematics Competition)
Solution: From the conditions, we have
$$
\begin{array}{l}
x y+(x+y)=71, \\
x y(x+y)=880,
\end{array}
$$

Therefore, $x y$ and $x+y$ are the two roots of the equation
$$
z^{2}-71 z+880=0
$$

Solving this equation, we get
$$
x+y=16, x y=55,
$$

or $x+y=55, x y=16$.
When $x+y=55, x y=16$, $x$ and $y$ are the two roots of the equation $z^{2}-55 z+16=0$. Solving this equation, we find that its roots are not positive integers, so this solution should be discarded. Similarly, we can verify that $x$ and $y$ satisfying $x+y=16, x y=55$ meet the conditions. Therefore, substituting 16 and 55 for $x+y$ and $x y$ respectively, we get
$$
\begin{array}{l}
x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=16^{2}-2 \times 55=256-110=146 .
\end{array}
$$