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!

Check this proof of Hensel's Lemma, which is basically the justification/proof of why what André did works, to go over the following:
$$x^2=2\pmod 7\iff f(x):=x^2-2=0\pmod 7\iff x=\pm 3\pmod 7$$
Choose now one of the roots, say $\;r=3\;$ , and let us "lift" it to a solution $\;\pmod{7^2}\;$ (check this is possible since $\;f'(3)=6\neq 0\pmod 7\;$)
$$s:=3+\left(-\frac77\cdot 6^{-1}\right)\cdot=3-(-1)^{-1}\cdot7=3+7=10$$
and indeed
$$f(10)=10^2-2=98=0\pmod{49=7^2}$$
Again:
$$s=10+\left(-\frac{98}7\cdot20^{-1}\right)\cdot7=10+14\cdot22\cdot7=2,166=108\pmod{7^3=343}$$
and indeed
$$f(108)=108^2-2=11,662=0\pmod{7^3}$$
Of course, you can repeat the above with every root in each stage.
