Let $E$ be an elliptic curve over some field $k$ of characteristic not $2$. Let $m$ be the maximal number of $4$-torsion points of $E(L)$ where $L$ ranges over all finite field extensions of $k$ of odd degree. What is the best lower bound on $m$ that does not depend on $k$ and $E$?
I can at least show that one always has $m\geq 2$. Indeed, over the separable closure of $k$ we have four $2$-torsion points but one of those is the neutral element. If there was no other $2$-torsion point over $k$, then we can get (at least) one more by passing to a field extension of degree $3$.
On the other hand, there are examples when $m\leq 4$. Indeed, if $k=\mathbb{R}$ and $E(\mathbb{R})$ has only one connected component, then there are exactly four $4$-torsion points and $\mathbb{R}$ has no odd field extension.
Best Answer
Your lower bound $m \geq 2$ is the best possible. In fact, I claim that if $E$ has at least one nontrivial $2$-torsion point $P$ over $k$, then all $2$-power torsion points of $E$ are defined over extensions of $k$ of $2$-power degree. In particular, if the $2$-part of $E(k)_{\mathrm{tors}}$ is just $\mathbb Z/2\mathbb Z$ (and of course there are many such $E$), then there are no additional $2$-power torsion points over any odd-degree extensions of $k$.
Proof: fix $n \geq 1$, and choose a basis of $E[2^n] = (\mathbb Z/2^n \mathbb Z)^2$ in which $P = (2^{n-1}, 0)$. The mod-$2^n$ Galois representation of $E$ fixes $P$, so its image is contained in the subgroup $$ \begin{pmatrix} 1 + 2\mathbb Z/2^n \mathbb Z & * \\ 2\mathbb Z/2^n \mathbb Z & 1 + 2\mathbb Z/2^n \mathbb Z \end{pmatrix} \subset \mathrm{GL}_2(\mathbb Z/2^n \mathbb Z). $$ But this subgroup has $2$-power order--namely order $2^{4n-3}$. Therefore its kernel is $G_L < G_k$, where $L$ is a $2$-power extension of $k$. This $L$ is the smallest field over which all $2^n$-torsion points of $E$ are defined.