We can proceed from your idea, by setting $a=2^x$ and $b=3^x.$Then, consider the expression
$$S=a^2+a^3+b^2-1-a-ab-a^2b$$
which is a quadratic with respect to $b$ and the discriminant is $\Delta =(a+1)^2(a-2)^2$, therefore:
$\displaystyle{S=(b-a-1)(b-a^2+1)=(3^x-2^x-1)(3^x-4^x+1)=f(x)-1}$or$\displaystyle{f(x)-1=8^x\left[ \left(\frac{3}{2} \right)^x-\left(\frac{1}{2} \right)^x-1\right]\left[ \left(\frac{3}{4} \right)^x+\left(\frac{1}{4} \right)^x-1\right]}$
Now $\displaystyle{ g (x)=\left(\frac{3}{2} \right)^x-\left(\frac{1}{2} \right)^x-1}$ is obviously increasing with unique root 1 and
$\displaystyle{h(x)=\left(\frac{3}{4} \right)^x+\left(\frac{1}{4} \right)^x-1}$ is decreasing with unique root 1.
Therefore $f(x)-1$ has only one root, for $x=1$, and it is negative if $x\neq 1$.
