Hints:
$2n+1$ is odd.
$2^{k(2n+1)} + 1 = 2^{k(2n+1)} + 1 + 2^{2kn} - 2^{2kn}$
try to show that $2^k + 1 | (2^{k(2n+1)} + 1)$ if and only if $2^k + 1 | (2^{k(2n-1)} + 1)$
$\cdots$
$2^k + 1 | 2^k + 1 $ $\square$
EDIT:
Hint $2$ to Hint $3$:
$2^{k(2n+1)} + 1 = 2^{k(2n+1)} + 1 + 2^{2kn} - 2^{2kn} = 2^{2kn+k} + 1 + 2^{2kn} - 2^{2kn} =$ $=2^{2kn}2^k + 1 + 2^{2kn} - 2^{2kn} = 2^{2kn}(2^k + 1) + 1 - 2^{2kn} =$ Let $a = 2^{2kn}(2^k + 1)$, now the equality is:
$= a + 1 - 2^{2kn} = a + 1 - 2^{k(2n-1)+k} = a + 1 - 2^{k(2n-1)}2^k =$ $= a + 1 - 2^{k(2n-1)}2^k +2^{k(2n-1)} - 2^{k(2n-1)}= a + 1 - 2^{k(2n-1)}(2^k +1) +2^{k(2n-1)}$
Now let $b= 2^{k(2n-1)}(2^k +1)$, and you are left with:
$2^{k(2n+1)} + 1 = (a - b) +( 2^{k(2n-1)} + 1)$
Note that $a,b$ are divisible by $2^k +1$ to reach Hint $3$ (I read in some other post that you haven't learned modular arithmetic yet - once you do this becomes so so so so much easier!).
Now it smells like induction. Apply Hint $1$ and draw a square.