Example 2 (1) If $9^{x}-3^{x}-6=0$, find the value of $x$;
(2) If $4^{x}-6 \cdot 2^{x}+8 \leqslant 0$, find the range of values for $x$.
(1) Let $3^{x}=t$, then $t^{2}-t-6=0, t_{1}=-2$ (discard) $t_{2}=3$. Therefore, $x=1$.
(2) Let $2^{x}=t$, then $t^{2}-6 t+8 \leqslant 0,2 \leqslant t \leqslant 4$. Therefore, $1 \leqslant x \leqslant 2$.