Does there exist $n$ such that $2^{2^n} + 1$ is divisible by $9$? Struggling with this question, from a problem sheet:

Does there exist $n$ such that $2^{2^n} + 1$ is divisible by $9$?

I think the answer is no, but I can't prove it. Any hints?
 A: Note that $2^{2^{n+1}}=(2^{2^n})^2$. Is there any square $\equiv -1\pmod 9$?
A: Since $\phi(9)=6$, $2^m+1$ repeats every $6$ terms:
$$2^m+1=\{\color{#8080F0}{\overset{0}{2}},\overset{1}{3},\overset{2}{5},\color{#C00000}{\overset{3}{0}},\overset{4}{8},\overset{5}{6},\color{#8080F0}{\overset{6}{2}},\overset{7}{3},\overset{8}{5},\color{#C00000}{\overset{9}{0}},\dots\}$$
Thus, $9\mid2^m+1\implies m\equiv3\pmod{6}$. This means that $3\mid m$.
If $m=2^n$, then $3\nmid m$, and therefore, $9\nmid2^m+1=2^{2^n}+1$.
A: I use my "memory-resident formula" ;-) $$\{2^t-1,3\} = [t:2](1 + \{t,3\}) \tag 1 $$
where (1) the notation $k_p= \{a,p\} $ means the exponent, to which the prime $p$ is factor in the canonical primefactorization of $a$ such that for instance if  $a=2^{k_2}\cdot 3^{k_3} \cdots  p^{k_p} \cdots \to \{a,p\}=k_p $ , and (2) the expression $[a:b]$ is $1$ if $b$ divides $a$ and is $0$ if not. (I find it useful to have such formulae for a handful of small primes in my mind) 
Then, because $ 2^t+1 = {2^{2t}-1 \over 2^t-1 } $ we have
$$\begin{array}{}
\{2^t+1,3\} &=& \{2^{2t}-1,3\} &- \{2^t-1,3\} \\
            &=& [2t:2](1 + \{2t,3\}) &-[t:2](1 + \{t,3\}) \\
            &=& 1 + \{2t,3\}  &-[t:2](1 + \{t,3\}) \\
            &=& 1   -[t:2]  &+ \{2t,3\} -[t:2]\{t,3\}) \\
 \end{array}$$
Now because $  \{2t,3\}=\{t,3\}$ ( having the same number of primfactors $3$ ) we have also
$$\begin{array}{}
\{2^t+1,3\} &=& 1   -[t:2]  &+ (1 -[t:2]) \cdot \{t,3\}) \\
\{2^t+1,3\} &=& (1   -[t:2]) & \cdot  (1 +\{t,3\}) \\
 \end{array} \tag 2$$
Now if we apply this general formula to your problem, we ask for $t=2^n$ and this gives
$$\begin{array}{}
 k &=& \{2^{2^n}+1,3\} &=& (1   -[2^n:2])   (1 +\{2^n,3\}) \\
 & & &=& 1   -[2^n:2]   \\
 \end{array} \tag 3$$
and this is zero, if $n \gt 0$ and $1$ if $n=0$, so the primefactor $3$ can occur only if $n=0$ and its exponent is then $1$ so this is never divisible by $9=3^2$.        
(Remark: the "memory-resident-formula" can also be found using the keyword "LTE" ("lift the exponent" or "lower the exponent" - don't know exactly) but I'd developed that formalism without knowledge of that introduced concept and have made my own notations)
