There is no hope for this in any subcanonical topology coarser than the fppf topology, or more generally, any subcanonical topology in which morphisms $\operatorname{Spec} C \to \operatorname{Spec} K$ are not automatically covers when $K$ is a field and $C$ is non-trivial. But it is true with the fpqc topology.
Indeed, if $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$ is any morphism that is not a cover and $F$ is the sheaf image, then $F$ is not the top subsheaf of $\operatorname{Spec} K$; if we further assume that $A$ is not trivial then $F$ is also not the bottom subsheaf of $\operatorname{Spec} K$. (We have $\alpha \in F (A)$ if and only if there exist a cover of $\operatorname{Spec} A$ by affines $\operatorname{Spec} \beta_i : \operatorname{Spec} B_i \to \operatorname{Spec} A$ such that each $\beta_i \circ \alpha : K \to B_i$ factors through $\phi : K \to C$. So $\textrm{id}_K \in F (K)$ if and only if $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$ is a cover.)
Conversely, by the above argument, in any subcanonical topology such that $\operatorname{Spec} K$ only has the top and bottom subsheaves, it must be that every morphism $\operatorname{Spec} C \to \operatorname{Spec} K$ is either a cover or has $C$ trivial. Nothing in the argument assumes that $K$ is a field, but if we assume that covers are faithfully flat and $K$ is non-trivial, it will follow that $K$ is a field: because $K$ is non-trivial, there exist a field $L$ and a ring homomorphism $K \to L$, and $\operatorname{Spec} L \to \operatorname{Spec} K$ is a cover so $K \to L$ is faithfully flat, hence $K$ is an integral domain with a unique prime ideal, i.e. a field. Thus, with the fpqc topology, $\operatorname{Spec} K$ has only the top and bottom subsheaves if and only if $K$ is a field.