Using Hensel's Lemma to find the number of solutions. I am trying to determine the number of solutions of the congruence
$x^2 ≡ 1 \mod 2^k$ when $k \ge 3$
The statement of Hensel's Lemma that I use is the following:
Attempt: $x \equiv 1,3,5,7 \mod 8$, and for all of these $\tau = 1$ so by (i) since we have 2 distinct solutions $\mod 4$ we have two distinct solutions $\mod 8$? This is where I get lost.

The following is not really an answer to the question, since after the first sentence it does not mention Hensel's lemma.
Let $k\ge 3$. We are trying to solve $x^2\equiv 1\pmod{2^k}$. For concreteness let $2^k=1024$. We want to solve the congruence
$$(x-1)(x+1)\equiv 0\pmod{1024}.$$
So $x-1$ and $x+1$ will need to be consecutive even integers. 
Of any two consecutive even integers, one is congruent to $2$ modulo $4$, and the other is congruent to $0$ modulo $4$.  The one congruent to $2$ modulo $4$ has only one $2$ to contribute to the product $(x-1)(x+1)$.  So the other one must supply the rest of the $2$'s.
Thus $(x-1)(x+1)$ is divisible by $1024$ if and only if one of $x-1$ and $x+1$ is divisible by $512$.
This gives the four solutions $x\equiv 1\pmod{1024}$, $x\equiv 513\pmod{1024}$, $x\equiv 1023\pmod{1024}$ and $x\equiv 511\pmod{1024}$.
