Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that
\begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*}
Define $a = x + y$ and $b = xy$. Then $a + b = 71$ and $ab = 880$. Solving these two equations yields a [quadratic](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation): $a^2 - 71a + 880 = 0$, which [factors](https://artofproblemsolving.com/wiki/index.php/Factor) to $(a - 16)(a - 55) = 0$. Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$. For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa). In the second case, since all factors of $16$ must be $\le 16$, no two factors of $16$ can sum greater than $32$, and so there are no integral solutions for $(x,y)$. The solution is $5^2 + 11^2 = \boxed{146}$.