(x, y)

As $\sqrt{2016} = 12\sqrt{14}$ we must have $x = 14a^{2}, y = 14b^{2}$ with $a + b = 12, 0 \le a,b \le 12$ .

As there are $13$ possible integral pairs $(a,b)$ , there will also be $\boxed{13}$ ordered integral pairs $(x,y)$ .