Solution:
We have $x^{2}+y = y^{2}+x$ which can be written as $(x-y)(x+y-1)=0$.

The case $x = y$ yields $x^{2}+x-12=0$, hence $(x, y) = (3, 3)$ or $(-4, -4)$.

The case $y = 1-x$ yields $x^{2}+1-x-12 = x^{2}-x-11=0$ which has solutions $x = \frac{1 \pm \sqrt{1+44}}{2} = \frac{1 \pm 3\sqrt{5}}{2}$. The other two solutions follow.