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?

$$\begin{align*}
2^{3k+2}+3&\equiv 2^{3k+2}-4\\
&\equiv 4\left(2^{3k}-1\right)\\
&\equiv 4\left((2^3)^k-1\right)\\
&\equiv 4\left(1^k-1\right)\\
&\equiv 4\cdot0\\
&\equiv 0 \pmod 7\;,
\end{align*}$$
since $8\equiv1\pmod7$.
