Does $n \mid 2^{2^n+1}+1$ imply $n \mid 2^{2^{2^n+1}+1}+1$?

My computer found the counterexample $n=520809$.

Interestingly $520809=57\times9137$, but I couldn't find any neat "explanation" for that, nor a natural way that $520809$ would appear.

Other few counterexample are $2343441, 15622617, 15622617...$ Indeed, one can show that there are infinitely many of then since if $n$ works, $f(n)$ also works (since $a|b \Leftrightarrow f(a)|f(b)$)

In case anyone is interested in my python code:

def factor(n): #Factors n, and returns a list with its factors (possibly repeated) in increasing order
 k=2
 v=[]
 while n>1:
  if n%k==0:
   v.append(k)
   n=n/k
  else:
   k=k+1
  if k*k>n:
   v.append(n)
   return v


def phi(n): #Calculates the totient function using the prime factorization
 v=factor(n)
 prod=1
 a=1
 for b in v:
  if b==a:
   prod*=b
  else:
   prod*=b-1
   a=b
 return prod 

def f2(n): # Calculates f(f(n)) mod n
 a=pow(2,n,phi(n))+1
 return (pow(2,a,n)+1)%n

def f3(n): # Calculates f(f(f(n))) mod n
 a=pow(2,n,phi(phi(n)))+1
 b=pow(2,a,phi(n))+1
 return (pow(2,b,n)+1)%n


n=11
while 1: #Tries all odd n
 if f2(n)==0:
  if f3(n)!=0:
   print n
   break
 n=n+2