Your equation is equivalent to the following statement: an element $x\in{\mathbb Z}_p$ is congruent to $1\pmod{p^{k+1}}$ if and only if one can write $x=y^{p^k}$, where $y\in{\mathbb Z}_p$ is congruent to $1\pmod p$. I believe you can prove this via Hensel's lemma, starting by looking $\pmod{p^{k+1}}$.
The analogous statement for $p=2$ can, I believe, be proved similarly; it is probably the same statement except with $k+1$ on the right-hand side replaced by $k+2$.
This is related to the fact that the multiplicative groups $({\mathbb Z}/p^k{\mathbb Z})^\times$ are cyclic when $p$ is odd, but $({\mathbb Z}/2^k{\mathbb Z})^\times$ isn't quite: it's the direct product of $\{\pm 1 \bmod 2^k\}$ with a cyclic group. This can be found in any number theory textbook that discusses primitive roots.