jmc

We are given that $x^2-y^2=133$, which is the same as $(x+y)(x-y)=133$. $133$ has two pairs of factors: 1 and 133, and 7 and 19. So, either $x+y=133$ and $x-y=1$, or $x+y=19$ and $x-y=7$. It is clear that $x$ and $y$ will be much larger in the first case, because they must sum to 133, so, because we are trying to minimize $x^2+y^2$, we can simply consider the second case. Through simple algebra, we find that $x=13$ and $y=6$. Thus, $x^2+y^2$ is minimized as $169+36=\boxed{205}$.