Square roots modulo powers of $2$
When $n\ge 3$, the number of solutions of $x^2\equiv 1\pmod{2^n}$ is $4$. The solutions are $x=\pm 1\pmod{2^n}$ and $x\equiv \pm 1+2^{n-1}\pmod{2^n}$.
Proof: We want $x^2-1\equiv 1\pmod{2^n}$, that is, $(x-1)(x+1)\equiv 0\pmod{2^n}$. Since $x$ must be odd, the gcd of $x-1$ and $x+1$ is $2$. Either all of the $2$'s come from $x-1$, or all the $2$'s come from $x+1$, or $n-1$ of them come from one of $x-1$ or $x+1$, and $1$ of them comes from the other.
To connect this with square roots, note that $u$ and $v$ are square roots of $a$, then $uv^{-1}$ is a square root of $1$, and conversely. So if $a$ has a square root, it has $4$ of them.
To show all $a$ congruent to $1$ mod $8$ have a square root, we use a counting argument. Note that there are $2^{n-3}$ numbers between $1$ and $2^n-1$ that are congruent to $1$ modulo $8$, and $2^{n-1}$ odd numbers. Since the squaring function is $4$ to $1$, it must be the case that every number congruent to $1$ modulo $8$ is the square of something.
The modular equation $x^2\equiv1\pmod{2^n}$ can be written this way: $$(x+1)(x-1)\equiv0\pmod{2^n}$$ It is clear from here that $x+1$ and $x-1$ are even. Since the difference between $x+1$ and $x-1$ is $2$, one of these numbers is a multiple of $2$ but not of $4$. This implies that the other one must be multiple of $2^{n-1}$.
This gives only four possibilities: $x\equiv1$, $x\equiv-1$, $x\equiv2^{n-1}+1$ and $x\equiv2^{n-1}-1$, that are distinct because $n\geq3$ (all these congruences are modulo $2^n$).
If $n\ge3$ and $a=8k+1$ then $a$ has exactly four distinct square roots modulo $2^n$.
Proof that there is at least one square root: use induction. The case $n=3$ is easy to check; if $x^2\equiv a\pmod{2^n}$ then $x^2=a+2^nk$ for some integer $k$; it is easy to see that $x$ is odd and so $$(x+2^{n-1}k)^2=a+2^nk(1+x)+2^{n+1}(k^22^{n-3})\equiv a\pmod{2^{n+1}}\ .$$
Finding all square roots: since $a\equiv x^2\pmod{2^n}$ we need to solve $y^2=x^2\pmod{2^n}$. This gives $$2^n\mid (y-x)(y+x)\ .$$ Each factor on the RHS is even; but not both are divisible by $4$ as then their sum $2y$ would be a multiple of $4$, which is impossible as $y$ is odd. So we have $$2^{n-1}\mid y-x\quad\hbox{or}\quad 2^{n-1}\mid y+x\ .$$ This gives the four possibilities $$y\equiv x\,,\ y\equiv x+2^{n-1}\,,\ y\equiv -x\,,\ y\equiv -x+2^{n-1}\pmod{2^n}\ ;$$ it is not hard to check that they are all different modulo $2^n$, and that they are in fact square roots of $a$.