Since the intersection $\mathbb Q(\zeta_m)\cap \mathbb Q(\zeta_n)=\mathbb Q(\zeta_{\gcd(m,n)})$ (for example here), it suffices to show that $\sqrt2$ is not contained in any cyclotomic field strictly inside $\mathbb Q(\zeta_8)$. The largest such field is $\mathbb Q(\zeta_4)=\mathbb Q(i)$, and it's clear that $\sqrt2$ is not in this field (since the only reals in $\mathbb Q(i)$ are the rationals), so the only $n$ for which $\sqrt2\in\mathbb Q(\zeta_n)$ are multiples of $8$.
For the more general case of $\sqrt{m}$ with $m=\pm p_1\cdots p_s$, we can use that, for an odd prime $p$,
$$\sqrt{(-1)^{(p-1)/2}p}\in\mathbb Q(\zeta_p).$$
(This can be proven, for example, by using Gauss sums.) From this, we definitely get that
$$\sqrt m\in\mathbb Q(\zeta_{4|m|}),$$
since $\sqrt{\pm p_i}$ is in this field for all $i$ (since if $2|m$, $8|4m$, this is true for $p=2$ as well). If $m$ is odd and of correct sign, i.e.
$$m=\prod_{i=1}^s (-1)^{\frac{p_i-1}{2}}p_i,$$
then we can reduce $4|m|$ to $|m|$.
We can prove that this in fact optimal by strong induction on $s$. For the base case, we need only worry about factors of $p$ or $4p$, and so the cases aren't that bad since $\sqrt{\pm p}$ isn't in $\mathbb Q(i)$ regardless of $p$ and $\mathbb Q(\zeta_n)=\mathbb Q(\zeta_{2n})$ for all odd $n$. For the inductive step, using any factor of $|m|$ or $4|m|$ either reduces the numbers of factors of $2$ (in which case arguments similar to the base case work) or removes a subset of the prime factors, in which case our strong inductive hypothesis can be used directly.
