For the first question, the two implications fail to reverse. For the first implication, just take $\mathfrak{n}$ to be a dually Dedekind infinite, Dedekind finite cardinal. For the second implication, see Theorem 3.5 of the paper Definitions of finite and the power set operation by Howard and Spišiak (which is not published, see this link for a copy).
For the second question, the answer is yes. See the paper Ein Beitrag zur Kardinalzahlarithmetik ohne Auswahlaxiom by Läuchli; for an English version, see Theorem 5.28 of the book Combinatorial Set Theory: With a Gentle Introduction to Forcing (second edition) by Halbeisen.
The key question in this field (proposed by Läuchli) is whether it is provable in $\mathsf{ZF}$ that $2^{2^{\mathfrak{n+1}}}=2^{2^{\mathfrak{n}}}$ (or equivalently, $(2^{2^{\mathfrak{n}}})^2=2^{2^{\mathfrak{n}}}$) for all infinite cardinals $\mathfrak{n}$. I conjecture the answer is no.