Trivial?

$x^2+y^2+2=(x-1)(y-1)$

$x^2+y^2=xy-x-y-1$

Multiplying both sides by 2

$2x^2+2y^2=2xy-2x-2y-2$

On rearranging this gives

$(x+1)^2+(y+1)^2+(x-y)^2=0$

This is only possible when all three terms are zero simultaneously

thus $x=y=-1$ is the only possible solution

$\quad x^2+y^2+2=(x-1)(y-1) \\ \quad x^2+y^2+x+y-xy+1 = 0 \\ \quad y^2+y(1-x)+(x^2+x+1)=0 \\ \quad \text{Discriminant Must be greater than or equal to } 0 \\ \quad D\ge 0 \implies (1-x)^2-4(x^2+x+1)\ge0 \\ \quad x^2+1-2x-4x^2-4x-4 \ge 0 \\ \quad -3x^2-6x-3 \ge 0 \implies x^2+2x+1\le 0 \\ \quad \boxed{ x = -1 } \\ \quad y^2+2y+1=0 \implies \boxed{ y = -1} \\ \quad \text{Only 1 real Solution exist for this equation } \boxed{(-1,-1)}$