Example 7 How many real numbers $x$ are there such that the value of the function $y=\frac{x^{2}-2 x+4}{x^{2}-3 x+3}$ is an integer?
Solve the equation $(1-y) x^{2}+(3 y-2) x+4-3 y=0$ for $x$. If $y=1$, then $x=-1$;

If $y \neq 1$, then $\Delta=(3 y-2)^{2}-4(1-y)(4-3 y) \geqslant 0$, which means $3 y^{2}-16 y+12 \leqslant 0$, then $\frac{8-2 \sqrt{7}}{3} \leqslant y \leqslant \frac{8+2 \sqrt{7}}{3}$, where $y \neq 1$ is satisfied by $y=2,3,4$.
When $y=2$, $x=2 \pm \sqrt{2} ; y=3$ 时, $x=1$ or $x=\frac{5}{2} ; y=4$ 时, $x=2$ or $x=\frac{4}{3}$.
Therefore, there are seven real numbers $x$ that make the value of the function $y=\frac{x^{2}-2 x+4}{x^{2}-3 x+3}$ an integer.

(Note: The last two sentences contain a mix of English and Chinese. Here is the corrected and fully translated version:)

When $y=2$, $x=2 \pm \sqrt{2} ; y=3$, $x=1$ or $x=\frac{5}{2} ; y=4$, $x=2$ or $x=\frac{4}{3}$.
Therefore, there are seven real numbers $x$ that make the value of the function $y=\frac{x^{2}-2 x+4}{x^{2}-3 x+3}$ an integer.