Let $X$ and $Y$ be two schemes over a base field $k$, and let $f:X\to Y$ be a morphism of schemes. I was wondering if the following condition is sufficient for $f$ to be a homeomorphism between the topological spaces of $X$ and $Y$
For every field extension $E/k$ we have $f(E): X(E) \to Y(E)$ is bijective.
More generally, what kind of information (if any) does the set $X(E)$ of $E$-points of $X$ give about the topological space of $X$?
What you lack is integrality.
Firstly, let me show that your criterion implies that $f$ is universally injective and bijective:
It is clear that your condition implies that $f$ is surjective, thus universally surjective by 01S1. Your condition also implies that $f$ is injective on points and a point on $X$ has the same residue field as its image in $Y$. Thus by 01L6, we know that $f$ is a monomorphism, in particular, $f$ is universally injective, see this answer.
Now 04DF shows that universal homeomorphism is equivalent to integral and universally injective and bijective.
You also asked about what if $f$ is of finite type. Integral and locally of finite type implies finite 01WJ. So to find an $f$ that satisfies your condition but not a homeomorphism, you should look for non-finite morphisms. Indeed, there are a lot of them: for example, take $0\coprod \mathbb{G}_m\to \mathbb{A}^1.$