$4^{\left(x^{2}+2\right)}-9 \cdot 2^{\left(x^{2}+2\right)}+8=0$
$\Rightarrow \left(2^{\left(x^{2}+2\right)}\right)^{2}-9 \cdot 2^{\left(x^{2}+2\right)}+8=0$
Put $2^{\left(x^{2}+2\right)}=y$, then $y^{2}-9 y+8=0$ which gives $y=8$ and $y=1$.
When $y=8$, then $2^{x^{2}+2}=8$
$\Rightarrow 2^{x^{2}+2}=2^{3}$
$\Rightarrow x^{2}+2=3$
$\Rightarrow x^{2}=1$
$\Rightarrow x=1,-1$
When $y=1$, then $2^{x^{2}+2}=1$
$\Rightarrow 2^{x^{2}+2}=2^{0}$
$\Rightarrow x^{2}+2=0$
$\Rightarrow x^{2}=-2$
which is not possible.