Four, (Full marks 16 points) Given that $x$ and $y$ are positive integers, and they satisfy the conditions $x y + x + y = 71$, $x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$.

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$$
\begin{array}{l}
(x+1)(y+1)=72, \\
x y(x+y)=2^{4} \times 5 \times 11 .
\end{array}
$$

From equation (1), we know that at least one of $x$ and $y$ is odd. Without loss of generality, let $x$ be odd.
From equation (2), $x$ can only be 5, 11, or 55.
However, when $x=55$, $x+y \geqslant 56, x y(x+y) \geqslant 55 \times 1 \times$ $56>880$, so $x=5$ or 11.
If $x=5$, from equation (1) we know $y=11$;
If $x=11$, we know $y=5$, in this case the values of $x$ and $y$ satisfy equation (2).
Therefore, $x^{2}+y^{2}=5^{2}+11^{2}=146$.