Denote $a = xy-x-y$. Then: \begin{align*} n=(x+1)^2-xy(2x-xy+2y)+(y+1)^2 = (xy-x-y)^2-2(xy-x-y-1) = a^2-2a+2 \end{align*}Since $a \geq -1$ There is only one such $a$. So we must find minimum $a$ such that number of $(x,y)$ that satisfy $a=xy-x-y$ is $61$. Since $a+1 = (x-1)(y-1)$, that means we must find minimum number that has $61$ divisors. Which is $2^{60}$. So minimum $n$ is equals to $n=(2^{60}-2)^2+1=\boxed{2^{120}-2^{62}+5}$ @below Thanks, fixed

EmilXM wrote: Denote $a = xy-x-y$. Then: \begin{align*} n=(x+1)^2-xy(2x-xy+2y)+(y+1)^2 = (xy-x-y)^2-2(xy-x-y-1) = a^2-2a+2 \end{align*}Since $a \geq -1$ There is only one such $a$. So we must find minimum $a$ such that number of $(x,y)$ that satisfy $a=xy-x-y$ is $61$. Since $a+1 = (x-1)(y-1)$, that means we must find minimum number that has $61$ divisors. Which is $2^{60}$. So minimum $n$ is equals to $n=(2^{60}-1)^2+1=\boxed{2^{120}-2^{61}+2}$ Typo above: since $a=2^{60}-1$, it should be $n=(a-1)^2+1=\left(2^{60}-2\right)^2+1$.