How to compute 2-adic square roots?
Because the derivative of $x^2-17$, i.e. $2x$ is $0 \bmod{2}$ Hensel's Lemma doesn't work very cleanly. In this situation when going from $p$ to $p^2$ either there is no lift, or every lift will work$\bmod p^2$. Let's look at what happens here -
$x^2\equiv 17 \bmod 2 \text{ has the solution }x\equiv 1 \bmod 2$
$(2y+1)^2 \equiv 17 \bmod 4 \text { is always true, telling us } x\equiv 1,3 \bmod 4 \text{ both work}$
When we lift to$\bmod 8$ we find $1$ and $5$ (lifts of $1 \bmod 4\,$) both work$\bmod 8$ as well as $3$ and $7$ (the lifts of $3 \bmod 4$). Note that we seem to have 4 solutions! Let's look at$\bmod 16$ and beyond. $$ \begin{array}\\ 1,5\pmod 8 & 1^2 \equiv (1+16) \equiv 17 \pmod{16} & 5^2\equiv 9 \not \equiv 17 \pmod{16} \\ 3,7\pmod{ 8} & 3^2 \equiv 9 \not\equiv 17 \pmod{16} & 7^2\equiv 49 \equiv 17 \pmod{16} \\ \end{array} $$ So of our 4 solutions only $1$ and $7\bmod 8$ will lift to$\bmod 16$. We lift those and try$\bmod 32$. $$ \begin{array}\\ 1,9\pmod{16} & 1^2 \not\equiv 17 \pmod{32} & 9^2\equiv 81 \equiv 17 \pmod{32} \\ 7,15\pmod{16} & 7^2 \equiv 49 \equiv 17 \pmod{32} & 15^2\equiv 225 \not\equiv 17 \pmod{32} \\ \end{array} $$ So of our 4 solutions only $9$ and $7\bmod 16$ will lift to$\bmod 32$. We lift those and try$\bmod 64$. \begin{array}\\ 9,25\pmod{32} & 9^2 \equiv 81 \equiv 17 \pmod{64} & 25^2\equiv 625 \not\equiv 17 \pmod{64} \\ 7,23\pmod{32} & 7^2 \equiv 49 \not\equiv 17 \pmod{64} & 23^2\equiv 529 \equiv 17 \pmod{64}\end{array}
Fairly tedious stuff for humans, but nothing a computer algebra system won't whip out in no time. We have found 2 roots, $1 + 2^3 + O(2^5)$ and $1 + 2+ 2^2 + 2^4 + O(2^5)$.
When Doing the calculations by hand it would probably make more sense to find only one root and multiply by $-1=\frac{1}{1-2}=1+2+2^2+...$ for the other root.
The binomial formula $(1+x)^\frac 12 = 1 + \frac12 x - \frac 18 x^2 + \ldots$ converges if $x \equiv 0 \pmod 8$, which gives you a way to find square roots for any $y \equiv 1 \pmod 8$.
In fact, the squares in the $2$-adics are the direct product of $\langle 4 \rangle$ with $(1+8\Bbb Z_2)$.
Here you can apply this directly to $17$, and it will converge even faster since it's $1 \pmod {16}$
Another thing this tells you is that a square root of $1+x$ is close to $1+\frac x2$, so you can recursively compute it by saying $\sqrt{1+8x} = (1+4x)\sqrt{(1+8x)/(1+4x)^2} = (1+4x)\sqrt{1-16(x/(1+4x))^2)}$. This gives you an infinite product whose terms are closer and closer to $1$. The number of correct digits doubles on each iteration
One way to apply Hensel cleanly is to use it to find not $\sqrt{17}$ but $(1+\sqrt{17}\,)/2$, whose minimal polynomial is $X^2-X-4$. If you want to use Newton-Raphson instead of Hensel, that too works more cleanly on $X^2-X-4$.