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.

Here is yet another way in. Note that $y$ must be a factor of $30$ as must $x+y$ (both from the second equation).
You can rewrite the first as $(y^2+x)(y^2-x)=xy+11=$ and use the second $=41-y^2$
Now note also that changing the signs of both $x$ and $y$ still gives a solution, so we can assume $y\gt 0$. 
If both $x$ and $y$ are positive $xy+11\gt 0$ and $y\le 6$
If $y$ is positive and $x$ is negative, then $x+y\gt 0$ from the second equation so $y\gt -x$, and since $y^2 \ge y$ we have also that $y^2\gt -x$, whence $(y^2+x)(y^2-x)\gt 0$ and $y\le 6$
So positive $y$ is restricted to $y\le 6$ and the possibilities are easily tested.
