Let $k$ be a complete, non-archimedean field, and $X$ a Berkovich space over $k$ (as nice as you like, for arguments sake let's say strictly $k$-analytic, good, and geometrically connected). As discussed in this article of de Jong, covering spaces of $X$ come in two slightly different flavours. One the one hand you can take finite etale covers $Y\rightarrow X$ as you would for schemes, on the other hand you can take a covering space $Y\rightarrow |X|$ of the underlying topological space of $X$, and, roughly speaking, use the Berkovich space structure of $X$ to put one on $Y$. Following de Jong, let us call the first of these 'algebraic' and the second 'topological'. A general covering space is then some kind of mixture of the two.

If $k$ is not separably closed, then one way of producing algebraic covering spaces is via finite separable extensions of $k$: if $L/k$ is such an extension then $X_L \rightarrow X$ is a finite etale map of Berkovich spaces, where $X_L$ denotes the base change of $X$ to $L$. My question is then the following:

**Question**: Is it possible that $X_L \rightarrow X$ is a **topological** covering space, for some non-trivial extension $L/k$?

It's not to hard to see this can't be the case if $X$ has a $k$-rational point (since the fibre of $X_L\rightarrow X$ over this point will have cardinality 1), but I'm particularly interested in the case when we might have $X(k)=\emptyset$. Concretely, I'm interested in the case when $X$ is (the analyitification of) a smooth projective conic over $k$, without a rational point, and $L/k$ is a quadratic extension over which $X$ does admit a rational point.