Find all integer solutions to $p^n+n=(n+1)^k$ where p is a prime of the form $2^m+1$

Question

Find all integer solutions to : $$p^n+n=(n+1)^k,$$

where $p$ is a prime of the form $2^m+1$.

I tried using binomial expansion and substituting $n$ with $c\times 2^m$ where c is a real number... but this seems to get nowhere.

Answer 1

Not a full answer (Observations/Research):

• By Polynomial Remainder Theorem, we have that the RHS has remainder 1 on division by $n$
• By observation, the LHS has the same remainder as $p^n$ Which by previous statement, and equality, means it has remainder 1.
• By a consequence of Lagrange's theorem in group theory, we have that the order of $p \pmod n$ divides the order of the group $Z/nZ$.
• By Euler's totient theorem, we have the order of the group, divides $\varphi (n)$, implying via previous observations that $n\over \varphi (n)$ is integer.
• Assuming $n$ is prime for a moment, we get the LHS is $p\pmod n$, therefore implying $p\equiv 1\pmod n$ by previous observations. This further would imply that, $n$ divides $p-1$ which is a power of 2
• etc.