Infinitely many positive integers $n$ such that $n$ divides $2^{2^{n}+1}+1$ and $n$ doesn’t divide $2^{n}+1$ Show that there are infinitely many positive integers $n$ such that $n$ divides $2^{2^{n}+1}+1$ and $n$ does not divide $2^{n}+1$
I think we should increase $n$ by induction:
$x_{k+1}$=$2^{x_{k}}+1$, then we have $x_{k}$ doesn’t divide $x_{k+1}$ and divide $x_{k+2}$.
Next we should show that if it allows for $k=1,2,3,..m$, then allows for $k=m+1$.
I am not getting if this approach is right.
Thanks in advance.

I don't see any way to solve the problem using induction. Instead, with the Lifting-the-exponent lemma, using $x = 2$, $y = 1$ and $p = 3$, then since $3 \mid 2 + 1$, for any positive integer $j$ and odd integer $k$ where $3 \not\mid k$,
$$\nu_3(2^{k(3^j)} + 1) = \nu_3(2 + 1) + \nu_3(k(3^j)) = 1 + j \tag{1}\label{eq1A}$$
which means $3^j \mid 2^{k(3^j)} + 1$. Next, for any integer $j \ge 2$ and odd integer $k$,
$$2^9 + 1 = 513 = 19(3^3) \implies 19 \mid 2^{k(3^j)} + 1 \tag{2}\label{eq2A}$$
We also have
$$2^{19} + 1 = 524289 = 3(174763) \tag{3}\label{eq3A}$$
Using the multiplicative order, let
$$\operatorname{ord}_{174763}(2) = m \tag{4}\label{eq4A}$$
Since $2^{2(19)} \equiv 1 \pmod{174763}$, then $m \mid 2(19)$. As $m \neq 1, 2, 19$, then $m = 2(19)$. Thus, for a positive integer $q$ and some non-negative integer $r$,
$$2^q \equiv -1 \pmod{174763} \iff q = 19 + rm \tag{5}\label{eq5A}$$
i.e., $q$ is an odd integral multiple of $19$.
For any integer $s \ge 2$, with $n = 174763(3^s)$, then since $19 \not\mid n$ (actually, $174763$ is a prime number), \eqref{eq5A} indicates $174763 \not\mid 2^n + 1$, so $n \not\mid 2^n + 1$. However, \eqref{eq2A} shows $19 \mid 2^n + 1$, so \eqref{eq5A} gives that $174763 \mid 2^{2^{n} + 1} + 1$. Also, as shown in \eqref{eq1A}, $3^s \mid 2^{n} + 1$ so $3^s \mid 2^{2^{n} + 1} + 1$. Thus, $n \mid 2^{2^{n} + 1} + 1$. This shows there's an infinite number of $n$ which meet the question conditions.
