$$x^2+xy-2y=64$$$$y=\frac{64-x^2}{x-2}$$$$y=-x-2+\frac{60}{x-2}$$This means, $x-2$ is a divisor of $60$. This gives $$x-2=1,2,3,4,5,6,10,12,15,20,30,60\implies x=3,4,5,6,7,8,12,14,17,22,32,62$$Checking each of these one by one, the only positive integer solutions are $$\boxed{(x,y)=(3,55),(4,24),(5,13),(6,7)}$$

Suppose there exists positive integers $x$ and $y$ and nonnegative integer $k$, $$x^2+xy-2y=k^2$$and $x+y+2$ is prime. Case I. $k\le2$. Since $x>2$, this meanss $$xy-2y=k^2-x^2\le0$$Therefore, $y(x-2)\le0$. This is a contradiction, because we know that $y>0$ and $x>2$. Therefore, $k\le2$ cannot happen. Case II. $k>2$. We have: $$x^2+xy-2y=k^2\implies y=\frac{k^2-x^2}{x-2}\implies x+y+2=\frac{k^2-4}{x-2}\implies x+y+2=\frac{(k+2)(k-2)}{x-2}$$Since $\frac{(k+2)(k-2)}{x-2}$ must be a prime number, then either $k-2\mid x-2$ or $k+2\mid x-2$. This means, $k-2\le x-2\implies k\le x$ However, we know that $x>2$, so $x^2+xy-2y>x^2\implies k^2>x^2\implies k>x$. This is a contradiction. Therefore, there does not exist $(x,y,k)$ which make $x+y+2$ prime. Hence, $x+y+2$ must be composite.