Third degree on the left but second degree on the right

I really like this problem and here is my solution.

Since $x+y \neq 0$ , then

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

$\Rightarrow$

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

Getting the discriminant of the equation with variable $x$ , we have

$(y+1)^2 -4(y^2-y)$

Since we want our $x$ values to be real, the discriminant must be greater than or equal to zero,

$\Rightarrow$

$-3y^2 + 6y+1 \geq 0$

Solving this we get the solution as

$\frac{1}{3}(3-2\sqrt{2} \leq y \leq\frac{1}{3}(3+2\sqrt{2})$

or

$-0.154701 \leq y \leq 2.1547$

Testing integer values for $y$ for the given interval and by doing casework, we find that these are the integer solutions

$(0,0), (1,0), (0,1), (2,1), (1,2), (2,2)$

In total there are $\boxed{5}$ ordered pairs excluding $(0,0).$