Prove that there exist an infinite number of $n\in \mathbb{N}$ such that $\frac{2^{n}+1}{n}=k, k\in \mathbb{N}$ I am quite sure that Vieta jumping (not sure if this is quite the right name for what I am doing? It is a jumping of sorts) works for this one, but I cannot prove it. The obvious first solution is $n = 3$, from then on we jump to the solution $n = 2^{3}+1=9$ which also produces a valid solution ($\frac{2^{9}+1}{9}=57$), from then on it follows that $n=2^{513}+1$ is also a solution, and WolframAlpha is able to verify this. I cannot compute the following ones by hand or machine, because the numbers get absolutely humongous very fast, but I have a hunch that this will work towards infinity.
I tried to get a recursive formula for all such $n$:
$$f(n)=2^{f(n-1)}+1,$$
$$f(0)=3$$
but I am also unable to get a nice expression for any $f(n)$ without recursion.
I also tried to do a proof by induction:
Suppose that:
$$\frac{2^{n}+1}{n}=k, k\in \mathbb{N}$$
then prove, that:
$$\frac{2^{2^{n}+1}+1}{2^{n}+1}=j, j\in \mathbb{N}$$
But also didn't get very far. Am I even correct in my assumption that this works in general? If so, how could I prove this? Thank you!
P.S. I am quite aware of the much easier proof by induction when $n=3^{k},k\in \mathbb{N}$, but this is something I happened to notice and it caught my attention.
 A: So let's suppose we have positive integers $n,k$ such that $2^n + 1 = nk$. Then
\begin{align}
2^{2^n+1} + 1
&= 2^{nk} + 1 \\
&= (nk - 1)^k + 1 \\
&= \Biggl(\sum_{m = 0}^{k} \binom{k}{m}(nk)^{k-m}(-1)^{m}\Biggr) + 1 \\
&= \sum_{m = 0}^{k-1} \binom{k}{m}(nk)^{k-m}(-1)^m \tag{$k$ odd} \\
&= (nk)\cdot \sum_{m = 0}^{k-1} \binom{k}{m}(nk)^{k-m-1}(-1)^m \\
&= (2^n+1)\cdot \sum_{m = 0}^{k-1} \binom{k}{m}(nk)^{k-m-1}(-1)^m \\
&\equiv 0 \pmod{2^n+1}\,,
\end{align}
thus whenever $n \mid 2^n + 1$ then $2^n+1 \mid 2^{2^n+1}+1$.
A: $\frac{2^{2^{n}+1}+1}{2^{n}+1}= \frac {2^{nk} +1}{2^n+1}$
Let $a = 2^n$ then we have $\frac{a^k + 1}{a+1}$.  If $k$ is odd then $\frac{a^k + 1}{a+1}= (a+1)(a^{k-1} - a^{k-2} + ....... + a^2 - a + 1)$.
So if $k$ is odd then $\frac {2^{nk} +1}{2^n+1}$ is an integer.
And $nk = 2^n+1$ is odd.  So $k$ is odd too.
=====  previous answer below =====
Well.....
So $3$ is a solution and $2^3 + 1=9$ is a solution.  And as it turns out $2^9 + 1 = 513$ is a solution.
It sure looks like if $n$ is a solution, then $2^n+1$ will be one too.
So lets try by induction:
If $2^{n} + 1 = m$ and $\frac {2^n+1}n = \frac mn = k$ then
$2^m + 1 = 2^{nk} + 1 = (2^n)^k + 1 = (2^n+ 1-1)^k =(m-1)^k + 1$ 
And $(m-1)^k + 1\equiv 0 \pmod m$.
(Because $m-1\equiv -1\pmod m$ and $k = \frac mn = \frac {2^n+1}n$ is odd.)
That's it.  We're done.  If $n|2^n + 1$ then $(2^n+1)|2^{2^n+1} + 1$. 
