Let $R=\mathbb Z[\zeta_8]$. I conjectured that $R^\times$, which is the units of $R$, as a set, is $\{\sigma^n\zeta_8^m\mid n,m\in\mathbb Z\}$, where $\sigma=1+\sqrt2$. Hence, I can get $R^\times\cong\mathbb Z\times C_8$. Can we prove it?
Suppose $x=a+b\zeta_8+c\zeta_8^2+d\zeta_8^3$. Noticing the proof of $R$ being euclidean, or taking the denominator of $x^{-1}$, we get a norm $$N:R\to\mathbb N, x\mapsto (a^2+c^2)^2+(b^2+d^2)^2+4(ab+cd)(ad-bc).$$ As it can be verified that $N$ preserves multiplication, the question boils down to finding the integer solution of $$(a^2+c^2)^2+(b^2+d^2)^2+4(ab+cd)(ad-bc)=1.$$ I'm not sure how to continue. Considering $\operatorname{im} N$ being non-negative, I believe that there is a way to complete the square or quartic power. I tried factorization but I cannot get a quartic factorization.
On the other hand, if I'm given $R^\times\cong\mathbb Z\times C_8$, by noting that $\sigma=1+\sqrt2$ is a unit, and that the multiplicative subgroup generated by $\sigma$ and $\zeta_8$ is isomorphic to $\mathbb Z\times C_8$, we just need to verify all real units to find all units. This can be easily done by solving a Pell equation.
