$x^2+7=8y+xy=y(8+x)$ $8+x|x^2+7=x^2-64+71 \to x+8|71 \to x=63,y=56$

y={x^2+7}/{x+8} y=x-8+{71/x+8} 71/x+8 must be integer =>x+8=1,71 x=63,as x is positive x+8 cant be 1 =>x=63,y=56 is the only positive solution

$$x(x-y)=8y-7 \implies x^2-xy-8y+7=0 \implies y^2+32y-28=(y+16)^2-284=k^2 \implies y=56 \implies x=63$$