Equation with integers $x$, $y$ If $x$, $y$ positive integers ($x<y$), how can I solve the equation $x+y=14\sqrt{xy-48}$ ?

Say $z=x+y\geq 0$ then $$ z^2 = 14^2(zx-x^2)-14^2\cdot 48$$ so $z= 14t$ and thus we have $$t^2= 14tx-x^2-48$$ or $$t^2-14tx +49x^2 =48(x^2-1)$$ so $$(t-7x)^2= 48(x^2-1)$$ and now we have $t-7x = 12s$ for some integer $s$. So we have $$ 3s^2= x^2-1$$ If $3|x-1$ then $$x-1 =3u^2\;\;\;{\rm and} \;\;\;x+1 = v^2$$ where $u,v$ are relatively prime. In this case take mod 3 and we get $v^2 =_3 2$ which is impossible.  
If $3|x+1$ then $$x+1 =3u^2\;\;\;{\rm and} \;\;\;x-1 = v^2$$ where $u,v$ are relatively prime. Then we get Pell's equation $3u^2-v^2=2$  which has infinitely solutions...
