Inspired by Jerry Han Jia Tao

Subtracting the given equations we get: $x^3-y^3-x^2y+y^2x+x^2-y^2=0$ $\Rightarrow (x-y)(x^2+y^2+x+y)=0$ $\color{darkviolet}{1. ~x=y}$ $\text{Substituting x=y in the first equation gives }\\~x=\pm 1 ~and ~corresponding ~y=\pm 1.\\$ $\color{darkviolet}{2.~x^2+y^2+x+y=0}$ Substituting x+y=t in this equationto get xy= $\dfrac{t^2+t}{2}$ . Now adding the equations given in the question, we get $(x+y)(x^2+y^2-xy)+2-xy(x+y)-(x^2+y^2)=0$ Substituting x+y=t and using $x^2+y^2=-x-y(\color{darkviolet}{as~in~1})$ $\large t^3+3t^2-t-2=0$ $\Rightarrow (t-1)(t+1)(t+2)=0$ $\Rightarrow t=x+y=-x^2-y^2=-1,-2~~(~1~rejected)$ Solving with $x^2+y^2+x+y=0,\\$ we get two more solutions viz(0,-1)(-1,0) Hence there are $4$ pairs of solutions i.e $\Large (1,1)(-1,-1)(0,-1)(-1,0)$

$(1)-(2)\Rightarrow (x-y)(x^2+y^2+x+y)=0$ Case 1:x=y $x=y\Rightarrow x^2-1=0\Leftrightarrow x=y=\pm 1\rightarrow 2 pairs$ Case 2: $\left\{\begin{matrix} x^2+y^2+x+y=0 (1)& & \\ x^3+1-x^2y-y^2=0 (2)& & \end{matrix}\right.$ $(1)+(2)\Leftrightarrow x^3+x^2-x^2y+x+y+1=0$ $\Leftrightarrow (x+1)(x^2-xy+y+1)=0$ Case 2.1:x=-1 $\Leftrightarrow x=-1\rightarrow y^2+y=0\Leftrightarrow (x,y)=(-1,0);(-1,-1)$ So we get 1 more pair. Case 2.2: $\left\{\begin{matrix} x^2+y-xy+1=0 (3)& & \\ x^2+y^2+x+y=0 (4)& & \end{matrix}\right.$ $(3)-(4)\Rightarrow y^2+x+xy-1=0\Leftrightarrow (y+1)(x+y-1)=0$ case 2.2.1: $y=-1\rightarrow x^2+x=0\Rightarrow (x,y)=(0,-1)$ -1 more pair case 2.2.2: $x+y=1\rightarrow y^2+y+1=0\rightarrow (x,y)= \theta$ (No solution) So we have 4 pairs satisfy the system of equation