This occurs iff the field has characteristic 0. By KConrad's comment, being characteristic 0 is certainly a necessary condition. Conversely, given an algebraically closed field K of characteristic 0, we can use Zorn's Lemma to find a maximal ordered subfield F. Since K is algebraically closed, F must be real closed. But also K must be algebraic over F or else we could pick a transcendental element and adjoin it to F (make it infinitely larger than all elements of F, i.e., use lexicographic ordering). Hence K must be of degree 2 over F and thus F is a maximal proper subfield.

Also by KConrad's comment, this is never unique; just apply an automorphism of K that takes, e.g., $\sqrt[3]{2}$ to $\omega \sqrt[3]{2}$, where $\omega$ is a primitive cube root of unity.