$x^{x^2} + x^{x^8} =?$ Given $x^{x^4} = 4$ If x is any complex number such that  $x^{x^4} = 4$ , then find all the possible values of :
$x^{x^2} + x^{x^8}$
First, I used laws of exponents to give $18$ as answer. However , I realised that I've misused it. Further I used logarithms which yielded
$$ x^{x^2} + x^{x^8} = 4^{1/x^2} + 4^{x^2} $$
After this, I am stuck and can't proceed further. By hit and trial, $x = \sqrt{2}$ seems to be one of the solution. Any help is appreciated.

This answer only works (based on the original question) on the set of real numbers.
Hint:
\begin{align}
x^{x^4}=4&\implies \left(x^4\right)^{x^4}=4^4\\
&\implies x^4=4\\
&\implies x=\pm\sqrt 2=\pm 2^{\frac 12}.\end{align}

Justification about the step $$\left(x^4\right)^{x^4}=4^4\implies x^4=4$$
We know that if $x\ge0$ and $x,y\in\mathbb R$ then the equation $ye^y=x$ has exactly one solution $W_0(x)$. Hence, we have
$$x^4\ln x^4=\ln x^4 e^{\ln x^4}=4\ln 4\ge 0$$
So, the fact $x^4=4$ is an only possible solution.
