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The short answer is that not too much is known about this situation. Here are a couple of very , beyond the easy first observations that I will now list. I will call $D$ irreducible if it has the property you are interested in, i.e., every (commutative) subfield that properly contains the center is a maximal subfield.

1. If $D$ has prime degree, then $D$ is irreducible. This is obvious because every subfield is contained in a maximal subfield, and the maximal subfields all have the same dimension over $F$.
2. If the degree of $D$ has at least two prime factors, then $D$ is reducible. In this case you can factor $D$ as a tensor product of two division algebras of relatively prime degrees. Then you just take a maximal subfield in one of the two factors. This reduces us to considering algebras $D$ whose degree is a prime power.
3. If $D$ has composite degree and $D$ is a crossed product, then $D$ is reducible. Recall that $D$ is a crossed product if it has a maximal subfield $L$ that is Galois over $F$. So suppose that the Galois group is $G$, necessarily of composite order. Then there is a nonzero proper subgroup of $G$, hence $D$ is reducible. (This deduction sounds foolish, because the theorem that something is a crossed product is much stronger than what you are asking about. But asking if something is a crossed product is a standard question, so in this way you can connect your question to standard results.)
4. As a consequence of #3, every $D$ of degree 4 is reducible, and every $D$ of degree 8 and exponent 2 is reducible. That is because such algebras are cross products under $Z/2 \times Z/2$ (Albert) and $Z/2 \times Z/2 \times Z/2$ (Rowen) respectively.
5. If $D$ has degree $p^r > p$ for some $p$ prime and it happens that every finite extension of $F$ has dimension a power of $p$, then $D$ is reducible. Indeed, by Galois theory every maximal subfield contains proper subfields.

So the first open case is where $D$ has degree 8 and exponent at least 4 in the Brauer group, and the base field has extensions of degree not a power of 2.

## Translation in terms of algebraic groups

Your question is closely related to the question of whether the group $SL_1(D)$ has nonzero, proper connected subgroups. Well, $SL_1(D)$ always contains maximal tori. So the question is: Are there others? (If your field has nonzero characteristic, probably one should only consider reductive subgroups.) Subfields of $D$ correspond to tori in $SL_1(D)$, so your question is the same as asking: For what $D$ are maximal tori the only nonzero, proper reductive subgroups of $SL_1(D)$?

These sorts of questions are addressed in my joint paper with Philippe Gille Algebraic groups with few subgroups, J. London Math. Soc., vol 80 (2009), 405-430. http://dx.doi.org/10.1112/jlms/jdp030 See especially section 4.

Also, the paper Irreducible tori in semisimple groups by Gopal Prasad and Andrei Rapinchuk (IMRN 2001, #23, 1229-1242) http://ams.rice.edu/leavingmsn?url=http://dx.doi.org/10.1155/S1073792801000587 discusses a similar question for tori. Your maximal subfield has no proper intermediate fields if and only if the corresponding torus is irreducible in their sense. This is why I called your $D$ irreducible above.

1

The short answer is that not too much is known about this situation. Here are a couple of very easy first observations. I will call $D$ irreducible if it has the property you are interested in, i.e., every (commutative) subfield that properly contains the center is a maximal subfield.

1. If $D$ has prime degree, then $D$ is irreducible. This is obvious because every subfield is contained in a maximal subfield, and the maximal subfields all have the same dimension over $F$.
2. If the degree of $D$ has at least two prime factors, then $D$ is reducible. In this case you can factor $D$ as a tensor product of two division algebras of relatively prime degrees. Then you just take a maximal subfield in one of the two factors. This reduces us to considering algebras $D$ whose degree is a prime power.
3. If $D$ has composite degree and $D$ is a crossed product, then $D$ is reducible. Recall that $D$ is a crossed product if it has a maximal subfield $L$ that is Galois over $F$. So suppose that the Galois group is $G$, necessarily of composite order. Then there is a nonzero proper subgroup of $G$, hence $D$ is reducible.
4. As a consequence of #3, every $D$ of degree 4 is reducible, and every $D$ of degree 8 and exponent 2 is reducible. That is because such algebras are cross products under $Z/2 \times Z/2$ (Albert) and $Z/2 \times Z/2 \times Z/2$ respectively.
5. If $D$ has degree $p^r > p$ for some $p$ prime and it happens that every finite extension of $F$ has dimension a power of $p$, then $D$ is reducible. Indeed, by Galois theory every maximal subfield contains proper subfields.

So the first open case is where $D$ has degree 8 and exponent at least 4 in the Brauer group, and the base field has extensions of degree not a power of 2.

## Translation in terms of algebraic groups

Your question is closely related to the question of whether the group $SL_1(D)$ has nonzero, proper connected subgroups. Well, $SL_1(D)$ always contains maximal tori. So the question is: Are there others? (If your field has nonzero characteristic, probably one should only consider reductive subgroups.) Subfields of $D$ correspond to tori in $SL_1(D)$, so your question is the same as asking: For what $D$ are maximal tori the only nonzero, proper reductive subgroups of $SL_1(D)$?

These sorts of questions are addressed in my joint paper with Philippe Gille Algebraic groups with few subgroups, J. London Math. Soc., vol 80 (2009), 405-430. http://dx.doi.org/10.1112/jlms/jdp030 See especially section 4.

Also, the paper Irreducible tori in semisimple groups by Gopal Prasad and Andrei Rapinchuk (IMRN 2001, #23, 1229-1242) http://ams.rice.edu/leavingmsn?url=http://dx.doi.org/10.1155/S1073792801000587 discusses a similar question for tori. Your maximal subfield has no proper intermediate fields if and only if the corresponding torus is irreducible in their sense. This is why I called your $D$ irreducible above.