$\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2$
$\Rightarrow \quad \log _{2}\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=2 \log _{2} 2$
$\Rightarrow \quad \log _{2}\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=\log _{2} 2^{2}$
$\Rightarrow \quad\left(\frac{9^{x-1}+7}{3^{x-1}+1}\right)=4$
$\Rightarrow \quad \frac{\left(3^{2}\right)^{x-1}+7}{\left(3^{x-1}+1\right)}=4$
$\Rightarrow \quad \frac{\left(3^{x-1}\right)^{2}+7}{3^{x-1}+1}=4$
Let $\quad 3^{\mathrm{x}-1}=\mathrm{y}$
$\therefore \quad \frac{y^{2}+7}{y+1}=4$
$\Rightarrow \quad \mathrm{y}^{2}+7=4 \mathrm{y}+4$
$\Rightarrow \quad y^{2}-4 y+3=0$
$\Rightarrow \quad y^{2}-3 y-y+3=0$
$\Rightarrow y(y-3)-1(y-3)=0$
$\Rightarrow \quad(\mathrm{y}-3)(\mathrm{y}-1)=0$
$\Rightarrow \quad y=3,1$
If $y=3$, then
$3^{x-1}=3$
$\Rightarrow \quad x-1=1 \Rightarrow x=2$
If $\mathrm{y}=1$, then
$3^{x-1}=3^{0}$
$\Rightarrow \quad x-1=0 \Rightarrow x=1$
$\therefore \quad x=1,2$