I need to show that $1+2^{(2^n)}+2^{(2^{n+1})}$ is divisible by $7$, for $n\in\mathbb{N}$.
Letting $x=2^{(2^n)}$,the expression becomes $1+x+x^2$.
Now, $1+x+x^2=\frac{x^3-1}{x-1}$.
We have $x^3=(2^{(2^n)})^{3}=8^{2^n}\equiv1 \pmod 7$.
So, $7\mid x^3-1$.
How to proceed now?
Looking at the remainders of $2^{(2^n)}$ when divided by $7$,for $n=0,1,2,3,4,....$,we get the pattern
$2,4,2,4,2,4,....$ , which can be proved by a simple induction.
So the original expression $1+2^{(2^n)}+2^{(2^{n+1})}$ considered modulo 7, i.e,
$[1+2^{(2^n)}+2^{(2^{n+1})}] \text{ mod } 7$ is either equal to $(1+2+4) \text { mod } 7$ or $(1+4+2) \text { mod } 7$.We see that in both cases,it is 0 .
$\therefore 7\mid 1+2^{(2^n)}+2^{(2^{n+1})}$.
How to proceed now?
Completing your original approach (in the question), $7\mid x^3-1=(x-1)(x^2+x+1)$,
and since the remainder of $2^{2^n}$ when divided by $7$ is not $1$ (i.e., $7\nmid x-1$), as you found,
it follows by Euclid's lemma that $7\mid x^2+x+1$.
We will use Euler Theorem on Modulo that is for any coprime natural number $a, n$ $$a^{\varphi(n)} \equiv a \mod n$$ Notice that for any natural number $a$ with $(a, 7) = 1$ $$a^6 \equiv 1 \mod 7$$ Notice that for any natural number $m$, $2^{2m} \equiv 4 \mod 6$ and $2^{2m+1} \equiv 2 \mod 6$. From there we can conclude that for any $n \in \mathbb{N}$, exist $k \in \mathbb{N}$ such that $\{2^{n}, 2^{n+1}\} = \{6k + 2, 6k + 4\}$ \begin{align*} 1 + 2^{2^n} + 2^{2^{n+1}} &\equiv 1 + 2^{6k + 2} + 2^{6k + 4} \mod 7 \\ & \equiv 1 + 2^2 + 2^4 \mod 7 \\ & \equiv 1 + 4 + 16 \mod 7 \\ & \equiv 21 \mod 7 \\ & \equiv 0 \mod7 \end{align*} Thus, for any $\displaystyle n \in \mathbb{N}: \ 7 \mid 1 + 2^{2^n} + 2^{2^{n+1}}$
