Substitution is definitely useful here

Relevant wiki: Wavy Curve Method

Given $x\in\mathbb Z~\cdots (1)$

Let's put $y=2^{|x|}~\cdots (2)$

Let $f(x) = \dfrac{(2^{|x|}+1)(2^{2|x|}+2^{|x|+2}+8)}{2^{|x|} (2^{2|x|}-16)(2^{|x|}-8)}$

$|x|\ge0\\\Rightarrow 2^{|x|}\ge 2^0=1\\\Rightarrow y\ge1\\\Rightarrow y\in\left[\left.-1,\infty\right.\right)~\cdots (3)$

So, $f(x) = g(y) = \frac{(y+1)(y^2+4y+8)}{y (y^2-16)(y-8)}$

We need $g(y)\ge 0$ along with $(1), ~(2)$ and $(3)$

Roots of expressions in $g$ : $y+1\rightarrow -1\\y^2+4y+8\rightarrow \text{no real roots}\\y\rightarrow 0 \\ y^2-16\rightarrow -4,~4\\y-8\rightarrow 8\\ y\ne -4, 0, 4, 8$

Effective degree of $g$ is odd

Applying Wavy-Curve method ,

So, we get $y \in (-4, -1]\cup(0,4)\cup(8,\infty)$

Since $y\ge1$ , $y \in [1,4)\cup (8,\infty)$

Integral values of $y$ which are not satisfying are: $y \in \{4,5,6,7,8\}\\\Rightarrow 2^{|x|}\in \{4,5,6,7,8\}, x\in \mathbb Z\\\Rightarrow 2^{|x|} \in \{4,8\}\\\boxed{x \in \{-3, -2, 2, 3\}}\\ \boxed{4~ \mathrm{ values}}$