Regarding the axiom $2^\kappa = 2^{\kappa^+}$ for regular cardinals $\kappa$, and its relationship to a couple of other axioms. (Take ZFC as background.)
The following two statements both follow from GCH:

ICF. Injective continuum function.
The continuum function (i.e. $\kappa \mapsto 2^\kappa)$ is injective.
NJA. No jumping axiom.
For all infinite cardinals $\kappa$ and $\nu$, we have: $$\kappa < 2^\nu \rightarrow 2^\kappa \leq 2^\nu.$$

To see that NJA follows from GCH, assume $\kappa < 2^\nu$. Hence $\kappa < \nu^+$. So $\kappa \leq \nu$. Thus $2^\kappa \leq 2^\nu$.
In fact, the converse holds too; we can actually prove GCH from the above two axioms. To see this, notice that GCH can be interpreted as saying that from $\kappa < 2^\nu$, we can derive $\kappa \leq \nu$. So assume $\kappa < 2^\nu$. Then from NJA, we deduce $2^\kappa \leq 2^\nu$. So from ICF, it follows that $\kappa \leq \nu$, as required.
Another interesting axiom that seems to be related to NJA is:

BA. Beth axiom.
For all infinite cardinals $\kappa$, there exists an ordinal $\alpha$ such that $2^\kappa = \beth_\alpha$.

I haven't been able to puzzle out whether or not NJA and BA imply each other, so:

Question 0. Is there a relationship between NJA and BA?

Okay. My motivation for factoring GCH as ICF+NJA is that I'm interested in axioms for set theory that determine the structure of the cardinal numbers (like GCH), but which don't trivialize the cardinal characteristics of the continuum (unlike GCH).
One approach to find such axioms is to look for statements that prove the truth of NJA, but which don't prove ICF. An obvious first axiom to consider is:

DCF. Degenerate continuum function.
For all infinite cardinal numbers $\kappa$, the following hold.
  
  
*
  
*If $\kappa$ is regular, then $2^\kappa = 2^{\kappa^+}$.
  
*$2^\kappa$ is regular.

This clearly refutes ICF. For example, under DCF, we can prove that $$2^{\aleph_0} = 2^{\aleph_1} = 2^{\aleph_2} = \cdots$$
and even that $$2^{\aleph_0} = 2^{\aleph_\omega}.$$
But, I haven't been able to puzzle out whether DCF implies either and/or both of NJA or BA.

Question 1. Does DCF imply either and/or both of NJA or BA?

 A: First of all, NJA implies BA. To see this, take a $\kappa$ and consider whether it is a beth number or not. If $\kappa=\beth_\alpha$ then of course $2^\kappa=\beth_{\alpha+1}$. Otherwise we can fit $\kappa$ between two beth numbers, $\beth_\alpha<\kappa<\beth_{\alpha+1}$, whence NJA implies that $2^\kappa=\beth_{\alpha+1}$.
On the other hand, the implication is not reversible: simply look at a model where $2^{\omega}=\omega_2, 2^{\omega_1}=2^{\omega_2}=\omega_3$ and GCH holds above. Here BA holds, since $2^{\omega_1}=\beth_2$, but NJA clearly fails at $\omega_1<2^{\omega}$.
For your second question, DCF does not imply BA (and therefore also not NJA). To get this, start with a model where $2^{\kappa}=\kappa^{+\omega+1}$ for every regular $\kappa$ (we can get this by Easton's theorem). 
This model is easily seen to satisfy DCF.
Now force over this model to get $2^\omega=\aleph_{\omega\cdot 2+1}$ and $2^{\aleph_{\omega+1}}=\aleph_{\omega\cdot 2+2}$. 
In the end we stil have DCF (we only changed exponentiation on two blocks of $\omega$ many cardinals) as well as $\beth_1=\aleph_{\omega\cdot 2+1}$ and $\beth_2=\aleph_{\omega\cdot 3+1}$. But if we look at $\kappa=\aleph_{\omega+1}$ the cardinal $2^\kappa=\aleph_{\omega\cdot 2+2}$ is not a beth number, so BA fails.
