Elegant way to factor?

Similar solution as @Alak Bhattacharya's

$\begin{aligned} (x^2+1)(y^2+1)+2(x-y)(1-xy) & = 4(1+xy) + 140 \\ x^2y^2 + x^2 + y^2 + 1 + 2x - 2y - 2x^2y + 2xy^2 -4xy & = 144 \\ x^2y^2 - 2x^2y + 2xy^2 +x^2 + y^2 -4xy + 2x - 2y - 143 & = 0 \end{aligned}$

Assume that the left-hand side of the equation above as:

$\begin{aligned} LHS & = (x^2 + ax + b)(y^2 + cy+d) + k \\ & = x^2y^2 + cx^2y + axy^2 + dx^2 + by^2 + acxy + adx+bcy + bd + k \end{aligned}$

Equating the coefficients, we have $c=-2$ , $a=2$ , $d=1$ , and $b=1$ , then $ac = -4$ , $ad = 2$ , and $bc=-2$ are correct and $bd=1$ , $\implies k = -144$ . Now we have:

$\begin{aligned} (x^2 + 2x + 1)(y^2-2y+1) - 144 & = 0 \\ (x+1)^2(y-1)^2 & = 144 \\ (x+1)(y-1) & = 12 \end{aligned}$

Since $12=2^2\times 3$ , there are $(2+1)(1+1)=6$ solutions as follows:

$\begin{array} {lll} 1 \times 12 = 12 & \implies \red{x = 0}, y = 13 & \small \red{\text{Unacceptable}} \\ 2 \times 6 = 12 & \implies x = 1, y = 7 & \implies xy = 7 \\ 3 \times 4 = 12 & \implies x = 2, y = 5 & \implies xy = 10 \\ 4 \times 3 = 12 & \implies x = 3, y = 4 & \implies xy = 12 \\ 6 \times 2 = 12 & \implies x = 5, y = 3 & \implies xy = 15 \\ 12 \times 1 = 12 & \implies x = 11, y = 2 & \implies xy = 22 \end{array}$

Therefore the sum of all possible $xy$ is $7+10+12+15+22 = \boxed{66}$ .

The given expression can be simplified to $(x+1)^2(y-1)^2=144$ . Since $x, y$ are positive integers, therefore $(x+1)(y-1)=12$ . Considering all the factors of $12$ , we get the pairs $(x, y)$ as $(1,7), (2,5), (3,4), (5,3), (11,2)$ , yielding the values of the product $xy$ as $7,10,12,15,22$ , whose sum is $\boxed {66}$ .