Generally when you work with field extensions, you want to assume that everything you consider sits inside some big algebraically closed field. For example in number theory, you regard all the finite extensions of $\mathbb{Q}$ as subfields of $\mathbb{C}$.
Let's fix an algebraic closure $K$ of $\mathbb Z_2$. So we can take all the extension fields of $\mathbb Z_2$ that we consider to be subfields of $K$. For each $m \geq 1$, there is a unique subfield $F$ (not even up to isomorphism, I really mean unique) of $K$ for which $[F : \mathbb Z_2] = m$.
Proof: Let $F$ be a field with $[F : \mathbb Z_2] = m$. Then $F$ has $2^m$ elements. And because $F \setminus \{0\}$ is a group with $2^m-1$ elements, you have that every element of $F$ is a root of the polynomial $X^{2^m} - X$. But the roots of this last polynomial in any algebraically closed field containing $\mathbb Z_2$ are distinct, so the roots of $X^{2^m}-X$ are exactly the elements of $F$. So if we assume $F$ is a subfield of $K$, then $F$ literally consists of those elements in $K$ which are roots of the polynomial $X^{2^m}-X$.
So back to your problem. You might as well assume $GF(2^3)$ to be a subfield of some fixed algebraic closure $K$ of $\mathbb Z_2$. If you take any irreducible factor $g$ of $X^{2^3}-X$, and you look at a field $\mathbb Z_2[x]$ where $x$ is some root of $g$, and you consider this to be a subfield of the same algebraically closed field $K$ as the one containing $GF(2^3)$, then you are going to literally have $\mathbb Z_2[x]$ contained in $GF(2^3)$, because $GF(2^3)$ is exactly the set of $y \in K$ satisfying $y^{2^3} = y$.
