It is not possible to prove there are no finite maximal antichains, because there are.

Paul Howard proved that Łoś's theorem and the Boolean Prime Ideal theorem imply in conjunction the Axiom of Choice. However there is a model in which every ultrafilter is principal, so Łoś's theorem holds trivially and therefore does not imply the Boolean Prime Ideal theorem; and there are models of the Boolean Prime Ideal theorem where Łoś's theorem fails (since the axiom of choice fails there). 

Bell and Fremlin proved all sort of similar conjunctions related to Hahn-Banach, Krein-Milman, and other geometric theorems about Banach spaces, where the conjunction of two imply the axiom of choice. These constitute of finite maximal antichains as well.

And there are probably others, which are "less interesting", like "Every set can be injected into $\Bbb R\times\alpha$ for some ordinal $\alpha$" and "The real numbers can be well-ordered", which in conjunction imply the axiom of choice, but can hold separately without it.

Note that all the suggestions here are of size $2$. This is because any finite antichain can be reduced to one of size $2$; but we can always break principles up. We can replace "$\Bbb R$" by some other set which is the product of several distinct sets and then argue that only one of them can be well-ordered each time.

> **Edit:** Note that the above treats "antichain" in the Boolean algebraic sense. It is certainly not the case that any choice principle either implies/follows from Łoś's theorem or from the Boolean Prime Ideal theorem. If we are interested in comparability rather than compatibility, then the answer is that every maximal antichain is either trivial or [countably] infinite, as remarked by Emil below Joel's answer.

As for finite maximal chains, the situation here is a bit less interesting and a bit more strained. For a finite antichain to be maximal it means that given any cardinality $X$, $\sf AC_X$, must be compatible with one of the principles in the chain, or imply the axiom of choice in conjunction. And that means that somehow we are in a situation where you already have quite a lot of choice, and any amount of choice above a specific set will readily imply the axiom. Similarly for comparing cardinalities of sets and $\aleph$ numbers.

> **Edit:** As the algebra is in fact atomless, it follows that it is dense, so every finite chain can be extended by at least one more element which lies between two successive elements of the chain. Therefore every chain must be infinite.