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.

HINT.-If you want to have another way then you can do as follows:
$$y^2+xy-30=0\iff 2y=-x\pm\sqrt{x^2+120}$$ the only possibilities are $x=\pm1\text{ or } \pm7$ and $x=\pm1$ is not compatible. This proves that $(7,3)$ and $(-7,-3)$  are the only solutions.
