Find all integers $n>1$ such that$\frac{2^n+1}{n+1}$ is an integer. Find all integers $n>1$ such that  $\frac{2^n+1}{n+1}$ is an integer.

Second attempt : I'd like to know if it's correct or not. Thank you.
Since $\frac{2^n+1}{n+1}$ for all integers $n>1$ is equivalent to  $\frac{2^{n-1} \;+1}{n}$ for all integers $n>2$, 
we will find all integers $n>2$ such that $\frac{2^{n-1}+1}{n}$ is an integer instead.
Let $n=\displaystyle\prod_{i=1}^l p_i^{k_i}$, where $k_i \geq 1$, $p_i \in \text{prime}$.
Let $p_j$ be prime such that $p_j\mid n$ and $p_j$ has minimum $v_2(p_j-1)$.
Since $n \mid 2^{n-1}+1$, so $n\in$ odd, then $p_j\in$ odd.
so $v_2(p_j-1) \geq 1$, i.e.,  there exist $r_j, m_j \in \mathbb{N}$ and $m_j \in \text{odd}$ such that $p_j-1=2^{r_j}\cdot m_j$ 
we have $p_j \equiv 1 (\bmod{2^{r_j}})$
Since $v_2(p_j-1)$ is minimum and $n\in$ odd, so $n \equiv 1 (\bmod{2^{r_j}})$
then we can write $n$ in the form, $n = 2^mt+1$, where $m, t \in \mathbb{N}$ and $m\geq r_j$ 
so $n \equiv -1 (\bmod{2^m})$
$-1 \equiv 2^{n-1} \equiv 2^{ 2^m \cdot t}\equiv 2^{ 2^m\cdot t\cdot m_j} (\bmod{p_j})$, since $m_j$ is odd, 
$-1 \equiv 2^{(2^{r_j} \cdot m_j) 2^{m-r_j}\;\cdot t} (\bmod{p_j})$
$-1 \equiv 2^{(p_j-1)  2^{m-r_j}\;\cdot t} (\bmod{p_j})$
$-1 \equiv 1 (\bmod{p_j})$, then $p_j=2$, contradicts that $p_j\in$ odd.
Therefore, there is no such $n$.
 A: HINT:  The following procedure is a sort of descent proving the impossibility of solution.
It is obvious that the denominator cannot be even so $n$ should be even and the quotient (supposing it is integer) should be odd.
$$2^{2n}+1=(2n+1)(2k+1)\iff2^{2n}=4kn+2(k+n)\iff2^{2n-1}=2kn+k+n$$ It follows that $k$ and $n$ have the same parity. We have to consider the two possible cases which give 
$$►(1)\text{ both odd: } 2^{2n-1}==2^3k_1n_1+(2+1)(k_1+n_1)+2^2$$
$$►(2)\text{ both even: } 2^{2n-1}=2^3k_1n_1+2(k_1+n_1) $$
$(1)$ and $(2)$ give respectively $$2^{n-2}=2^2k_1n_1+(2+1)(k_1+n_1)+2\\2^{n-2}=2^2k_1n_1+k_1+n_1$$ 
Both cases require $k_1$ and $n_1$ have the same parity. 
Iterating the procedure we get in each step $k_i$ and $n_i$ have the same parity 
which leads, after sufficient iterations, to an equality in which an integer is equal to another greater integer. Contradiction. 
$$\text{ A concise example}$$ $$(2^6+1)=(6+1)(2k+1)=14k+6+1\iff2^6=14k+6\iff2^5=7k+3\\ 2^5=7k+3\Rightarrow2^5=7(2k_1+1)+3=14k_1+10\iff2^4=7k_1+5\\2^4=7k_1+5\Rightarrow2^4=7(2k_2+1)+5=14k_2+12\iff2^3=7k_2+6\text{
 absurde }.$$
