How Many Pairs?

Notice that $x\neq 0$ . Also notice that $y>0$ .

Let $x>0$ , since if $(x,y)$ is a solution, so is $(-x,y)$ . If we know all the positive solutions, we'll know all the negative ones.

Notice that $\begin{cases}x^2\equiv 0\pmod 3\\\text{or}\\x^2\equiv 1\pmod 3\end{cases}$ for all $x\in\mathbb Z$ . $\displaystyle x^2 + 615 \equiv x^2\equiv (-1)^y\pmod 3$

Hence $y$ is even. Let $y=2m$ , where $m\in\mathbb N$ .

$\displaystyle x^2+615=2^{2m}\iff (2^m+x)(2^m-x)=615=3\cdot5\cdot 41$

Since $2^m+x>2^m-x$ , the only possibilities are

$\displaystyle \begin{cases}\begin{cases}2^m+x=615\\2^m-x=1\end{cases}\\\text{or}\\\begin{cases}2^m+x=205\\2^m-x=3\end{cases}\\\text{or}\\\begin{cases}2^m+x=123\\2^m-x=5\end{cases}\\\text{or}\\\begin{cases}2^m+x=41\\2^m-x=15\end{cases}\end{cases}$

The only integer solution to any of the systems of equations above is $\begin{cases}x=59\\m=6\iff y=12\end{cases}$

Hence $(59,12)$ and $(-59,12)$ are the only integer solutions.

$\displaystyle \sum_{i=1}^{2}(-1)^ix_iy_i=-x_1y_1+x_2y_2=59\cdot12+59\cdot12=\boxed{1416}$