Question: Let $\mathfrak n$ be an infinite cardinal. In ZF (set theory without the axiom of choice) can either of the implications $$\mathfrak n=\mathfrak n+1\implies2^\mathfrak n=2^{\mathfrak n+1}\implies2^\mathfrak n=2^\mathfrak n+1$$ be reversed? Equivalently, can either of the implications $$\mathfrak n\ge\aleph_0\implies2^\mathfrak n=2^{\mathfrak n+1}\implies2^\mathfrak n\ge\aleph_0$$ be reversed?

Motivation: Idle curiosity.

What I know: The implications can't both be reversed because, if $\mathfrak n=2^\mathfrak m$ is incomparable with $\aleph_0$ (i.e., $2^\mathfrak m$ is a Dedekind-finite infinite cardinal) then $2^\mathfrak m\lt2^\mathfrak m+1$, while $2^{2^\mathfrak m}=2^{2^\mathfrak m}+1$ since $2^{2^\mathfrak m}\ge\aleph_0$ (Russell's theorem).

Question: Is $2^{2^\mathfrak m}=2^{2^\mathfrak m+1}$ if $\mathfrak m$ is incomparable with $\aleph_0$?

Question: Let $\mathfrak n$ be an infinite cardinal. In ZF (set theory without the axiom of choice) can either of the implications $$\mathfrak n=\mathfrak n+1\implies2^\mathfrak n=2^{\mathfrak n+1}\implies2^\mathfrak n=2^\mathfrak n+1$$ be reversed? Equivalently, can either of the implications $$\mathfrak n\ge\aleph_0\implies2^\mathfrak n=2^{\mathfrak n+1}\implies2^\mathfrak n\ge\aleph_0$$ be reversed?

Motivation: Idle curiosity.

What I know: The implications can't both be reversed because, if $\mathfrak n=2^\mathfrak m$ is incomparable with $\aleph_0$ (i.e., $2^\mathfrak m$ is a Dedekind-finite infinite cardinal) then $2^\mathfrak m\lt2^\mathfrak m+1$, while $2^{2^\mathfrak m}=2^{2^\mathfrak m}+1$ since $2^{2^\mathfrak m}\ge\aleph_0$ (Russell's theorem).

Question: Is $2^{2^\mathfrak m}=2^{2^\mathfrak m+1}$ if $2^\mathfrak m$ is incomparable with $\aleph_0$?