Example 23 (1988 Beijing Mathematics Competition) Find all ordered integer pairs $(x, y)$ that satisfy the equation $x^{2}+2 x y+3 y^{2}-2 x+y+1=0$.

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Solve: Transform the given equation into a quadratic equation in terms of $x$
$$
x^{2}+2(y-1) x+\left(3 y^{2}+y+1\right)=0 \text {. }
$$

If the equation has integer solutions, then a necessary condition is $\Delta=4\left[(y-1)^{2}-\left(3 y^{2}+y+1\right)\right] \geqslant 0$, from which we solve $-\frac{3}{2} \leqslant y \leqslant 0$. Since $y$ is an integer, so $y=0,-1$.
When $y=0$, $x^{2}-2 x+1=0, x=1$;
When $y=-1$, $x^{2}-4 x+3=0, x=1, x=3$.
Therefore, all ordered integer pairs that satisfy the given equation are
$$
(x, y)=(1,0),(1,-1),(3,-1) .
$$