Given an irreducible polynomial $p \in \mathbb{Q}[x]$ we're interested in how it factors after repeated simple field extension of degree $2$. So we generate a chain of fields $F_{n+1} / F_{n}$, where $F_0 = \mathbb{Q}$ and $F_{n+1} = F_{n}( \sqrt c_n)$, where $c_n \in F_n$
Let's assume $p$ factors over $F_n$ into non-constant polynomials, meaning $p = uv$, $u, v \in F_n[x]$. I can prove that if this is the first $n$ where $p$ is reducible, then $\deg(u)=deg(v)=\frac{\deg(p)}{2}$ must hold. It's not hard to get this by multiplying both sides of the equation by it's conjugate, which will yield $v=\overline{u}$.
This will mean that if $p$ factors, it can at first only do so into $2$ polynomials of degree $\frac{\deg(p) }{2}$, and later on the factors themselves can only split into polynomials half their own degree.
My question is whether such factorizations actually materialize for every irreducible polynomial, so can we pick a chain of field extensions, where over some $F_n$ we can write $p$ as a product of odd degree polyomials? I suspect the answer is yes, and if so it's also interesting how many extension do we need? I recon we should be able to factor at least one polynomial with every extension.
Consider the special case where the degree of $p$ is a power of $2$. If your conjecture were true, then the eventual odd-degree factors you would obtain would be linear, so $p$ would split over the resulting field. This means the splitting field of $p$ is a subfield of a field obtained by repeated quadratic extensions, so its degree over $\mathbb{Q}$ is a power of $2$. This then implies the order of the Galois group of $p$ is a power of $2$.
So, any irreducible polynomial whose degree is a power of $2$ but whose Galois group's order is not a power of $2$ is a counterexample. For instance, you could take a polynomial of degree $2^n$ (for $n>1$) whose Galois group is the full symmetric group $S_{2^n}$.