Example 4 Solve the equation $\left(4^{x}-6 \cdot 2^{x}\right)+\left(4^{-x}-6 \cdot 2^{-x}\right)+10=0$.
Solve the original equation, which can be rearranged as $\left(4^{x}+4^{-x}\right)-6\left(2^{x}+2^{-x}\right)+10=0$.
Let $t=2^{x}+2^{-x}$, then $4^{x}+4^{-x}=t^{2}-2$, the original equation becomes
$$
t^{2}-2-6 t+10=0,
$$

Solving for $t$ gives $t=2$ or $t=4$.
For $2^{x}+2^{-x}=2$, we have $\left(2^{x}\right)^{2}-2 \cdot 2^{x}+1=0$, which yields $2^{x}=1$, thus $x=0$;
For $2^{x}+2^{-x}=4$, we have $\left(2^{x}\right)^{2}-4 \cdot 2^{x}+1=0$, which yields $2^{x}=2 \pm \sqrt{3}$, thus $x=\log _{2}(2 \pm \sqrt{3})$.
In summary, the solutions to the original equation are $x_{1}=0, x_{2}=\log _{2}(2+\sqrt{3}), x_{3}=\log _{2}(2-\sqrt{3})$.