My gut feeling for solving this problem is to use strong induction.
Starting with the base case $n=1$ we can check each of the seven congruence classes and find that $x_1=2$ is the unique solution. Then assuming for $1\le m\le k$ there is a unique root modulo $7^m$. So we know from Hensel's Lemma we know that since there is a unique solution modulo $x_k$ to the congruence $f(x)\equiv 0\pmod{7^k}$ that $7$ does not divide $f'(x_{k-1})$. To show that when $n = k+1$ we have a unique solution it would be sufficient to show that $7$ does not divide $f'(x_k)$.
This is where I keep hitting a wall. I can't seem to figure out how to show that $7$ does not divide $f'(x_k)$. Using Hensel's Lemma Ive been able to find $x_k$ in terms of $x_{k-1}$ but its really ugly and involves inverses of $f'(x_{k-1})$.
Could I get a hint to try and steer me in the right direction, and if I'm going way off track what can I do to correct it?
You have that $f'(x)=3x^2+2x=x(3x+2)$. Mod $7$, the only roots to that are $\{0,4\}$. Could $x_k$ have been one of these mod $7$?
If $x_k\equiv0$ mod $7$, then $7$ does not divide $x_k^3+x_k^2-5$, so $x_k$ was not a root mod $7^k$.
If $x_k\equiv4$ mod $7$, then ... (leaving as a hint. See answer history for the full answer.)