2023 AMC 10B Problem 14

If $m = 0,$ the equation forces $n^2 = 0,$ giving $(0, 0).$ Otherwise both are nonzero; assume $|m| \le |n|.$ Then $m^2 n^2 = m^2 + mn + n^2 \le 3n^2,$ so $m^2 \le 3$ and $m = \pm 1.$ Take $m = 1:$ $1 + n + n^2 = n^2$ gives $n = -1.$ Take $m = -1:$ $n = 1.$ That leaves $(0,0), (1,-1), (-1,1),$ three in all. Therefore, the answer is C.