If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$. In this (btw, nice) answer to Twin primes of form $2^n+3$ and $2^n+5$, it was said that:

If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$?

I'm not familiar with these kind of calculations, so I'd like to see, if my answer is correct:


*

*Let $n=3k+2$ so then
$2^{3k+2}+3\equiv 2^{3k+2}-4 \equiv 4\left(2^{3k}-1\right)\equiv \phantom{4\cdot } 0 \pmod 7$.

*Reading a bit about Modular arithmetics, I found that 
$
a_1 \equiv b_1 \pmod N  $ and $  a_2 \equiv b_2 \pmod N 
$
then
$
a_1 a_2 \equiv b_1 b_2 \pmod n$, if $a_i$ and $b_i$ are integers.

*Since we have $4\equiv 4 \pmod 7$, I conclude that $2^{3k}-1 \equiv 0 \pmod 7$.

*Finally we use that
$
    2^{3n}-1=(2^3-1)\cdot \left(1+2^3+2^{2\cdot 3}+2^{3\cdot 3}+\cdots+2^{(k-1)3}\right)
$
and are done.
Are there alternative ways to prove it?
 A: $$\begin{align*}
2^{3k+2}+3&\equiv 2^{3k+2}-4\\
&\equiv 4\left(2^{3k}-1\right)\\
&\equiv 4\left((2^3)^k-1\right)\\
&\equiv 4\left(1^k-1\right)\\
&\equiv 4\cdot0\\
&\equiv 0 \pmod 7\;,
\end{align*}$$
since $8\equiv1\pmod7$.
A: In fact, we can prove a stronger result and the proof is easier. The result we will prove is that $$x^2+x+1 \text{ divides }x^{3k+2} + x+1$$ for all $k \in \mathbb{N}$. Setting $x=2$ gives the result, you are looking for.
The proof follows immediately from the remainder theorem since $(x^2+x+1) = (x-\omega)(x-\omega^2)$, where $\omega$ is the complex cube-root of unity.
(Remember that $(x-a)$ divides $f(x)$ if and only if $f(x) = 0$)
Plugging in $\omega$ in $x^{3k+2}+x+1$ gives us $\omega^{3k+2} + \omega + 1 = \omega^2 + \omega + 1 = 0$. This gives us that $(x- \omega)$ divides $x^{3k+2}+x+1$.
Similarly, plugging in $\omega^2$ in $x^{3k+2}+x+1$ gives us $\omega^{6k+4} + \omega^2 + 1 = \omega + \omega^2 + 1 = 0$. This gives us that $(x- \omega^2)$ divides $x^{3k+2}+x+1$.
Hence, $(x-\omega)(x-\omega^2) = x^2 + x + 1$ divides $x^{3k+2}+x+1$.
Setting $x=2$ gives us the result you want i.e. $2^2 + 2 + 1 = 7$ divides $2^{3k+2}+2+1 = 2^{3k+2}+3$.
EDIT (Deleted the other answer and merged with this)
Another way to prove $x^2+x+1 \text{ divides }x^{3k+2} + x+1$ for all $k \in \mathbb{N}$ is by induction. All that is needed for induction is that $$x^{3k+5} + x + 1 = x^3 \left(x^{3k+2} + x + 1\right) - x^4 - x^3 +x + 1\\
= x^3 \left(x^{3k+2} + x + 1\right) - x^3 (x+1) +x + 1\\
= x^3 \left(x^{3k+2} + x + 1\right) - (x^3-1) (x+1)$$
So if $x^2 + x +1$ divides $\left(x^{3k+2} + x + 1\right)$, then it also divides $x^{3k+5} + x + 1$ since $x^2 + x +1$ divides $x^3-1$. Also, the base case $k=0$ is trivially true since $x^2 + x + 1$ divides $x^{3 \times 0 + 2} + x + 1 = x^{2} + x + 1$.
Hence, $x^2 + x + 1$ divides $x^{3k+5} + x + 1$ forall $k \in \mathbb{N}$. Setting $x=2$ as before gives you what you want.
A: It doesn't require factorization or any other such ingenuity, just simple modular arithmetic:
$$\rm mod\ 7\!:\ \ 3 + 2^{\:\!2+3\:\!K} = 3 + 4\cdot 8^K \equiv 3 + 4\cdot 1^K\equiv 3+4 \equiv 0$$
