Solve $\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$ for $x$ 
Solve $$\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$$ for $x$ 

Any help is appreciated. I'm entering a challenge and can't reach the solution.

First, we can rearrange the given equation as follows:
\begin{align}\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}&=13\\\\
\frac{(2^2)^{2x}+(2^2)^x+1}{2^{2x}+2^x+1}&=13\\\\
(2^2)^{2x}+(2^2)^x+1&=13\left(2^{2x}+2^x+1\right)\\
2^{4x}+2^{2x}+1&=13\left(2^{2x}+2^x+1\right)\\
(2^x)^4+(2^x)^2+1&=13\left((2^x)^2+2^x+1\right)\\
y^4+y^2+1&=13\left(y^2+y+1\right)\tag{substitute $y:=2^x$}\\
y^4+y^2+1&=13y^2+13y+13\\
y^4-12y^2-13y-12&=0\end{align}
We are then left with a quartic equation to solve, and then we can solve $y=2^x$ for each of the values of $y$ we find. 
Can you continue from here?
