Solvability and reducibility of a polynomial in a "chain" of finite fields This question is generalized based on my previous question: Is $x^5 + x^3 + 1$ irreducible in $\mathbb{F}_{32}$ and $\mathbb{F}_8$?
Problem:
Consider an irreducible polynomial $f = x^4 + x^3 + 1$ in $\mathbb{F}_2 [x]$. I am interested in its solvability and reducibility in a "chain" of fields. Namely, does $f$ have root in the following fields? Is $f$ reducible in the following fields?
$$ \color{red}{ \mathbb{F}_2},\mathbb{F}_4,\mathbb{F}_8,\color{red}{\mathbb{F}_{16}},\mathbb{F}_{32},\cdots,\color{red}{\mathbb{F}_{256}}$$
My progress:
$\mathbb{F}_2,\mathbb{F}_{16},\mathbb{F}_{256}$:
It is easy to verify that $f$ has no root in $\mathbb{F}_2$ and it is irreducible as well. As a result, $f$ has a root in $\mathbb{F}_{16} \simeq \mathbb{F}_2/f(x)$ and thus is reducible in $\mathbb{F}_{16}$. Since $\mathbb{F}_{256} = \mathbb{F}_{2^{4\times 2}}$, $\mathbb{F}_{16}$ is a subfield of $\mathbb{F}_{256}$, $f$ has root and is reducible in $\mathbb{F}_{256}$ as well.

$\mathbb{F}_{32},\mathbb{F}_{64},\cdots$:
For the fields "between" $\mathbb{F}_{16}$ and $\mathbb{F}_{256}$, I verified that $f$ cannot have root in them. Take $\mathbb{F}_{64}$ for example. Suppose $f$ actually has root in it, call it $\zeta$. Then apparently $\mathbb{F}_2 (\zeta)$ is a subfield of $\mathbb{F}_{64}$. But since $\mathbb{F}_2 (\zeta) \simeq \mathbb{F}_2 [x]/f(x)$, this would imply that $\mathbb{F}_{16}$ is a subfield of $\mathbb{F}_{64}$, which is impossible. But I haven't figured out how to deal with the reducibility part?

$\mathbb{F}_4,\mathbb{F}_8$:
For the fields "between" $\mathbb{F}_2$ and $\mathbb{F}_{16}$, I am not sure what's the best strategy here. For the existence of root, I am attempted to say that $f$ has no root in both of them since "the smallest" field extension that has a root is $\mathbb{F}_{16}$. I forgot the proof but I think it is correct. 
With the operation tables of $\mathbb{F}_4$ from WolframAlpha, I was able to show that $f$ has no root in $\mathbb{F}_4$, and it is irreducible in $\mathbb{F}_4$ as well. But is there a better way?
For $\mathbb{F}_8$, I mimicked the answer in my previous question and managed to show that $f$ is irreducible in $\mathbb{F}_8$. But I don't know how I should deal with the existence of root?
I understand that this is a long post and I appreciate your help. Thanks.
 A: Here is a simple way to answer all these questions, just from the fact that $f$ is irreducible over $\mathbb{F}_2$.  Since $f$ is irreducible over $\mathbb{F}_2$, $\mathbb{F}_2[x]/(f)$ is a field.  Since $f$ has degree $4$, this field is $4$-dimensional over $\mathbb{F}_2$, and hence is isomorphic to $\mathbb{F}_{16}$.  But $f$ has a root in a field $K$ iff $\mathbb{F}_2[x]/(f)$ embeds in $K$ (by sending $x$ to the root in $K$).  This means $f$ has a root in $\mathbb{F}_{2^n}$ iff $4\mid n$.
In particular, then, an extension $\mathbb{F}_{2^m}$ of $\mathbb{F}_{2^n}$ will contain a root (and hence all the roots) of $f$ iff $4\mid m$.  Also, $\mathbb{F}_{2^m}$ is an extension of $\mathbb{F}_{2^n}$ iff $n\mid m$.  This means that adjoining any root of $f$ to $\mathbb{F}_{2^n}$ gives $\mathbb{F}_{2^{\operatorname{lcm}(4,n)}}$.  Thus each irreducible factor of $f$ over $\mathbb{F}_{2^n}$ must have degree $\frac{\operatorname{lcm}(4,n)}{n}=\frac{4}{\gcd(4,n)}$.  Explicitly, this means that $f$ has a root in $\mathbb{F}_{2^n}$ iff $4\mid n$, is reducible but has no root iff $2\mid n$ and $4\not\mid n$, and is irreducible iff $n$ is odd.
More generally, a similar analysis shows that if $f(x)\in\mathbb{F}_q[x]$ is irreducible of degree $d$, then over $\mathbb{F}_{q^n}$ $f$ splits as a product of irreducible factors of degree $\frac{d}{\gcd(d,n)}$.
