Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $ 
Find all pairs of intergers $(x,y)$ that satisfy the two following equations:
$x^2+11 = y^4 -xy$
$y^2 + xy= 30 $

Here's what I did:
$x^2+11 +(30) = y^4 -xy +(y^2 + xy)$
$x^2+41 = y^4  +y^2 $
$x^2 = y^4  +y^2 - 41$
$x^2 -49 = y^4  +y^2 - 41 - 49$
$(x+7)(x-7)= (y^2 +10) (y^2-9)$
And from here you get that two pairs can be: $(3,7)$ and $(-3,-7)$.
I think I've made some progress but don't know if those are the only two pairs that satisfy the equation.
Also I would like to see a different way of solving it since I think subtracting that $49$ was just luck.

Continuing your calculation,
$(\dfrac{30-y^2}y)^2=y^4+y^2-41,y≠0 $
$⇒y^4-60y^2+900=y^2(y^4+y^2-41)$
$⇒y^6+19y^2-900=0$
$⇒y^2(y^4+19)=900$
This give integer solution $y^2=9$.
