Let $F(x, y) = x^3 - x^2y + y^2 + x - y$. Note that $F(1, 1) = 1$ and $F(1, 2) = 2$. We shall prove that the equation $F(x, y) = 3$ has no solution in positive integers. Write this equation in the form
$$y^2 - (1 + x^2)y + x^3 + x - 3 = 0.$$
Its discriminant with respect to $y$ equals
$$D = (1 + x^2)^2 - 4(x^3 + x - 3) = x^4 - 4x^3 + 2x^2 - 4x + 13.$$
Since $D < (x^2 - 2x - 1)^2$ for any integer $x \ge 2$ and $D > (x^2 - 2x - 2)^2$ for any integer $x \ge 6$, the equation $F(x, y) = 3$ has no positive integer solutions if $x \ge 6$.
Direct verifications show the same if $x \in \{1, 2, 3, 4, 5\}$.
Hence the wanted integer is 3.