jmc

Let $s=x+y$ and $p=xy$. Then the first equation reads $s+p=71$, and the second equation reads $$x^{2}y+x{y}^2=(x+y)xy=sp=880.$$Therefore $s$ and $p$ are the roots of $$t^2-71t+880=0.$$This factors as $$(t-16)(t-55)=0,$$so $s$ and $p$ are the numbers $16$ and $55$ in some order. If $s=16$ and $p=55$, then $$x^2+y^2=(x+y{)}^2-2xy=s^2-2p=16^2-2\cdot 55=146.$$If $s=55$ and $p=16$, then from $x+y=55$, we see that $p=xy\ge 1\cdot 54=54$, which is a contradiction. Therefore the answer is $\boxed{146}$.