<s>[INST] Find all oddly even numbers $m$ such that $m^{2^n - 1} - 1 = (m - 1)(m^n + p^k)$ Find all natural numbers $m \equiv 2 \pmod{4}$ such that there exist exist two natural numbers $n, k$ and a prime number $p$ with $\dfrac{m^{2^{n}-1}-1}{m-1}=m^n+p^k$.
My solution:
The equation lead to $n \geq 2 $,hence $LHS \equiv3$  mod4.So,$p^k \equiv 3$ mod 4. Thus, $p \equiv 3$ mod 4. Then,
$m^{2^{n}-1}+(1-m)(m^{n+1}+1)= m(m-1)p^k$
Set $n+1=2^l(2h+1)$ hence $l<n$ and $m^{2^l}+1 \mid m^{2^n}-1$.We also have $$m^{2^l}+1 \mid m^{n+1}+1$$.Thus, $m^{2^l}+1 \mid m(m-1)p^k$. Because $\gcd \left( m(m-1), m^{2^l}+1 \right)=1$ so $m^{2^l}+1 \mid p^k$. 
If $l \ge 1$ then always exist a prime number $q \equiv 1 \pmod{4}$ and $ q|m^{2^l}+1$( a well-known problem), contradiction.
Thus,  $l=0$ and $m+1 \mid p^k$.So,  $m=p^l-1$ , $2 \nmid l$
We'll prove that for each number  $m=p^l-1$ there exist $n,k,p$.Really, we choose $n=2,k=l$
 [/INST] If $n=1$, $m+p^k=1$. Contradiction. 
Let $n\ge2$.
$$\frac{m^{2^n-1}-1}{m-1}=m^{2^n-2}+\ldots+m^2+m+1=m^n+p^k$$
$4$ divides every power of $m$ except $m$ thus $LHS\equiv m+1\equiv 3\pmod4$.
LHS is odd, and $m$ is even. So,$p$ is odd. If $2|k$, $p^k\equiv 1\pmod4$ and $4|m^n$. So,$RHS\equiv 1\pmod4$. Contradiction. So, $k$ is odd.
If $n=2$, we need $m+1=p^k$. There are infinitely many solutions with $m=p-1$, where $p$ is an arbitrary prime of the form $4k+3$, and $k$ is an arbitrary odd integer.
Suppose, $n\ge3$. Then, 
we have $p^k\equiv 3\pmod4$ and $p^k\equiv 1\pmod m$. If $d=\gcd(n,2^n-2)>1$, then, for any $q|\frac{m^{d}-1}{m-1}$, 
$$m^n\equiv 1\equiv \frac{m-1}{m-1}\equiv \frac{m^{2^n-2}m-1}{m-1}\pmod q$$
So, $q|p^k$, thus, $q=p$. So, $q<m^n$ and $q^k>m^{2^n-2}$, implies, $k>\frac{2^n-2}n$. In particular, $k\ne1$.
Moreover, $q=p$ is true for any divisor $q$ of ${m^d-1\over m-1}$, ${m^d-1\over m-1}$ should be a power of $q$. Which implies $d$ is a prime number by Zsigmondy's Theorem.(Exception for Zsigmondy: $m=2$,$d=6$ does not work here.)
Note: I'm aware I'm assuming something weird. But, this actually works for $n$ prime power or $\gcd(6,n)>1$. I tried to factorize $p$ and this looked like the pattern and unfortunately, the smallest $n$ that does not satisfy this property is $35$, which is out of my computer's ability to calculate.
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