You expect that when you try to take square root of $1+4u$, you're led to an unramified extension, just as ${\mathbb{Q}}_2(5^{1/2})$ is unramified over $ {\mathbb Q}_2$. Indeed, from $x^2 - (1+4u)$ you are led (by an appropriate change of variables) to $X^2 + X - u$, clearly either irreducible with roots in an unramified extension of your $2$-adic ground field, or reducible, depending on whether the corresponding polynomial in characteristic $2$ doesn't or does have roots in the residue field. That drops out of Hensel's Lemma.

The upshot is that you can ``always get a unit $u$ such that $1+4u$ is not a square'' if and only if the residue field has quadratic extensions. Always, in particular, if the residue field is finite.