Example 3 The number of solutions to the equation $x^{2}-2[x]-3=0$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
Solve: From $x^{2}-2[x]-3=0 \Rightarrow x^{2}-3=2[x]$.
And $[x] \leqslant x$, thus,
$$
\begin{array}{l}
x^{2}-3 \leqslant 2 x \\
\Rightarrow x^{2}-2 x-3 \leqslant 0 \\
\Rightarrow-1 \leqslant x \leqslant 3
\end{array}
$$
$\Rightarrow[x]$ possible values are $-1,0, 1,2,3$.
When $[x]=-1$, $x^{2}=1 \Rightarrow x=-1$;
When $[x]=0$, $x^{2}=3$, no solution that meets the condition;
When $[x]=1$, $x^{2}=5$, no solution that meets the condition;
When $[x]=2$, $x^{2}=7 \Rightarrow x=\sqrt{7}$;
When $[x]=3$, $x^{2}=9 \Rightarrow x=3$.
In summary, the original equation has 3 solutions, choose (C).