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

We have: $$x + y = 14\sqrt{xy -48} \implies 196(xy-48)= (x+y)^2 \implies x^2+y^2-194xy+9408 =0$$
Now, rearranging the terms we get: $$ y^2-194xy = -x^2-9408 \implies 9409x^2-194xy +y^2=9408x^2-9408$$ $$ \implies (y-97x)^2 = 9408(x^2-1) \implies y = 97x \pm 56\sqrt{3(x^2-1)}$$
There are many solutions to this: $(1,97), (2,26)$, etc.
