How did we find the solution? In my lecture notes, I read that "We know that $$x^2 \equiv 2 \pmod {7^3}$$ has as solution $$x \equiv 108 \pmod {7^3}$$"
How did we find this solution? 
Any help would be appreciated!

Without referring to Hensel I'd make the "Ansatz"
$$x=a\cdot 7^2+b\cdot7+c$$
with $|a|$, $|b|$, $|c|$ integers $\leq3$. This is no restriction of generality. The condition $x^2=2\ (7^3)$
leads to
$$(b^2+2ac)\cdot7^2+2bc\cdot7+ c^2-2=0\quad(7^3)\ .\tag{1}$$
It follows that $c^2=2\ (7)$, which implies $c=\pm3$. Plugging this into $(1)$ gives
$$(b^2\pm 6a)\cdot7^2 \pm 6b\cdot7+7=0\quad(7^3)\ ,$$
and dividing by $7$ leads to
$$(b^2\pm 6a)\cdot7+(\pm 6b+1)=0\quad(7^2)\ .\tag{2}$$
This implies $b=\pm1$. Inserting this into $(2)$ and dividing by $7$ again we obtain
$$(1\pm 6a)+1=0\quad(7)\ ,$$
and this is solved by $a=\pm2$. 
Therefore all solutions of the original equation have to be of the form
$$x=\pm(2\cdot7^2+1\cdot7+3)=\pm 108\quad(7^3)\ .$$
It is easily checked that $\pm108$ indeed does the job.
