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.