I think the answer is "hardly ever", because pretty much everything is algebraic maximal in your sense. For any complete discretely-valued field $K$, and any finite extension $L / K$, we have $[L : K] = e(L / K) f(L / K)$, where $f(L/K)$ is the degree of the extension of residue fields and $e(L / K)$ is the index of the value group of $K$ in that of $L$. So any complete discretely valued field is algebraic maximal, and such fields are very far from being algebraically closed!

I can't actually think offhand of an example of a valued field which is *not* algebraic maximal.