How many?

We have $x^2 - x = x (x-1) \geq 0$ and $y^2 \geq 0$ .

That limits the possibilities to $0 \leq y^2 \leq 3$ . Two cases:

-- $y = 0$ , but then $x^2 - x = 3$ , $(x - \tfrac12)^2 = 3\tfrac14$ , which has no rational solution;

-- $y = \pm 1$ , and then $(x-\tfrac12)^2 = 2\tfrac14$ , so that $x - \tfrac12 = \pm1\tfrac12$ and $x = -1$ or $x = 2$ . That means there are four solutions: $(-1, -1);\ (-1,1);\ (2, -1);\ (2, 1).$

$x^2-x+y^2-3=0$

The discriminant for $x$ in terms of $y$ is $13-4y^2$

A quick check tells us that $y^2$ can be only equal to $1$ to have a perfect square discriminant

So, $y=(1,-1)$ and since the original equation is a quadratic in $x$ , then for every value of $y$ there are $2$ values of $x$

Therefore, there are $2 \cdot 2 = \boxed{4}$ ordered pairs.

If one simply adds 1/4 to both sides of this Diophantine equation, we obtain the equation of a circle in the xy-plane:

x^2 - x + 1/4 + y^2 = 3 + 1/4;

or (x - 1/2)^2 + y^2 = 13/4 (i).

which is centered at (1/2, 0) and has radius = sqrt(13)/2.

This circle described in (i) has x-intercepts at ((1-sqrt(13)/2, 0) and ((1+sqrt(13)/2, 0), which has x = -1, 0, 1, and 2 as the only integral values belonging to this domain. Testing each value in (i) yields:

(x,y) = (-1, -1); (-1, 1); (0, sqrt(3)); (0, -sqrt(3)); (1, sqrt(3)); (1, -sqrt(3)); (2, -1); (2, 1).

Only four out of these eight ordered pairs are integral.