Let $R$ be an integral domain with $\operatorname{char}(R) = 0$ and let $\zeta, \zeta'$ be two primitive roots of unity in $R$. The following are equivalent.
(1) $(q-\zeta)^m \in (q-\zeta') + I[q]$ for some $m \geq 0$ and an ideal $I \subset R$ such that $R$ is $I$-adically separated (note: $(q-\zeta')$ on the right side denotes the ideal generated by $q- \zeta'$).
(2) $R$ is $(\zeta-\zeta')$-adically separated.
(3) $\operatorname{ord}(\zeta^{-1}\zeta') = p^k$ for some prime $p$ and $k \in \mathbb{Z}\setminus \{0\}$ such that $R$ is $(p)$-adically separated.
I believe I have just the implication (2) $\implies$ (1). If $R$ is $(\zeta-\zeta')$-adically separated, then $\cap_{j\geq 0}(\zeta-\zeta')^j = (0)$. We have to show that $(q-\zeta)^m \in (q-\zeta') + I[q]$ for some $m \geq 0$ and an ideal $I \subset R$ such that $R$ is $I$-adically separated. But $$q-\zeta \in (q-\zeta') + (\zeta-\zeta')[q]$$ since $q-\zeta = (q-\zeta') + (\zeta'-\zeta)$.
I sort of "have" the proof of the reciprocal, (1)$\implies$ (2), but I don't understand a step. The author says that since $(q-\zeta)^m \in (q-\zeta') + I[q]$ for some $m \geq 0$ and $I \subset R$ such that $R$ is $I$-adically separated, then $(\zeta-\zeta')^m \in I$. This is what I don't find obvious at all. From that I believe the result follows since the ideal generated by $(\zeta - \zeta')^m$ would be then contained in $I$, so $$\bigcap_{j\geq 0}(\zeta-\zeta')^j \subset \bigcap_{j\geq 0}I^j = (0)$$ and hence $R$ is $(\zeta-\zeta')$-adically separated.
For any of the implications (2)$\implies$(3) or (3)$\implies$(2) I literally have no clue. Any hero would be nice. Thank you!
First, note that the minimal polynomial (over $\Bbb Z$) of a primitive $p^k$th root of unity $\zeta$ is $$f(x) = \prod_{i\in\left(\Bbb Z/p^k\Bbb Z\right)^\times} (x - \zeta^i) = x^{p^{k-1}(p-1)} + x^{p^{k-1}(p-2)}+\cdots + x^{p^{k-1}} + 1.$$ Evaluating $f$ at $x = 1$ gives $\prod_i (1 - \zeta^i) = p.$
Now, for any fixed $i,$ $1 - \zeta^i = \epsilon_i(1 - \zeta),$ where $$\epsilon_i = \frac{1 - \zeta^i}{1 - \zeta} = 1 + \zeta + \dots + \zeta^{i - 1}\in R.$$ In fact, $\epsilon_i$ is a unit of $R$! Indeed, let $j$ be such that $ij\equiv 1\pmod{p^k}.$ Then we compute $$\epsilon_i^{-1} = \frac{1 - \zeta}{1 - \zeta^i} = \frac{1 - \zeta^{ij}}{1 - \zeta^i} = 1 + \zeta^{i} + \dots + \zeta^{i(j - 1)}\in R.$$ Thus, at the level of ideals we have $$ (p) = \prod_i (1 - \zeta^i) = \prod_i (1 - \zeta) = (1 - \zeta)^{\varphi(p^k)}. $$
More refined arguments may be found in most books on algebraic number theory; for example, in chapter 1 section 10 of Neukirch's Algebraic Number Theory.
For the implication (2)$\implies$(3), we note that we simply must show that $\xi = \zeta^{-1}\zeta'$ is a primitive $p^k$th root of unity. Indeed, once we have this, the previous argument showing $(1 - \zeta_{p^k})^{\varphi(p^k)} = (p)$ proves that $R$ is $p$-adically separated.
We will prove the contrapositive: we will assume that $\xi$ is a primitive $n$th root of unity, where $n$ is not a prime power, and prove that $R$ is not $(1 - \xi) = (\zeta - \zeta')$-adically separated. In fact, we shall prove that $(1 - \xi) = R.$
Let $p$ be any prime dividing $n.$ Then $\xi^r = \zeta_p$ is a primitive $p$th root of unity for some $r$, and $$ \frac{1 - \zeta_p}{1 - \xi} = \frac{1 - \xi^r}{1 - \xi} = 1 + \xi + \dots + \xi^{r - 1}, $$ so that $1 - \zeta_p\in(1 - \xi).$
Now, because $(1 - \zeta_p)^{p - 1} = (p),$ it follows that $p\in (1 - \xi).$ If $\ell$ is another prime dividing $n,$ the same argument shows that $\ell\in (1 - \xi)$ as well. Elementary number theory tells us that there exist integers $a$ and $b$ such that $$ 1 = ap + b\ell. $$ But writing $p = u(1 - \xi)$ and $\ell = v(1 - \xi)$ for some $u,v\in R,$ it then follows that \begin{align*} 1 &= ap + b\ell\\ &= au(1 - \xi) + bv(1 - \xi)\\ &= (au + bv)(1 - \xi)\\ \implies 1&\in (1 - \xi). \end{align*}
So, if multiple primes divide $\operatorname{ord}(\xi),$ we have $(1 - \xi) = R,$ and $$ \bigcap_i (1 - \xi)^i = \bigcap_i R = R, $$ which implies $R$ is not $(1 - \xi) = (\zeta - \zeta')$-adically separated (as $R$ is not the zero ring).
While I'm at it, I'll try to clear up the confusion about the argument provided that shows (1)$\implies$(2).
This is shown by simply evaluating at $q = \zeta'.$ Indeed, $(q−\zeta)^m\in(q−\zeta')+I[q]$ means precisely that $$ (q - \zeta)^m = P(q)(q - \zeta') + Q(q), $$ where $Q(q)\in I[q]$ and $P(q)\in R[q].$ Notice that $(q - \zeta)^m = \pm (\zeta - q)^m,$ depending on whether $m$ is even or odd, so that $$ (\zeta - q)^m = \tilde{P}(q)(q - \zeta') + \tilde{Q}(q), $$ where $\tilde{P}\in R[q]$ and $\tilde{Q}\in I[q].$ Setting $q = \zeta',$ we find $$ (\zeta - \zeta')^m = \tilde{P}(\zeta')(\zeta' - \zeta') + \tilde{Q}(\zeta') = \tilde{Q}(\zeta')\in I, $$ because each coefficient of $\tilde{Q}$ is in $I.$ The rest of your reasoning showing that $R$ is then $(\zeta - \zeta')$-adically separated is correct.