Show that $(k+1) \nmid 2^{k}+1$ for any $k\in \mathbb{N}$
I had found out this question from somewhere and here's my solution. I am not sure if I am missing out something in the solution. Please check if it's correct.
My partial solution (not sure if it adds any value) :
If $k$ is odd there's a clear contradiction since an even number can't divide an odd number, which shows $k$ is even.
This shows that $k+1$ is odd, meaning, it has got all odd prime divisors. Let $p$ be an odd prime dividing it. Since $k$ is even, observe that $2^{k}$ is a squared number and $2^{k}\equiv -1 \pmod{p}$. By the theorem stated here we see that $p = 4x+1$ for some $x$. This shows that each and every odd prime factor of $k+1$ is of the form $4j+1$ and so is $k+1$.
And I couldn't proceed anymore!