 Solution

For $z=a+b\sqrt{d}$, let $\bar{z}=a-b\sqrt{d}$ denote the algebraic conjugate, not necessarily ordinary complex conjugation when $d>0$. Conjugation respects multiplication, and $z\bar{z}=a^2-d{b}^2$. ThereforeIf $z$ is a unit in a ring, choose its inverse $w$ in the same ring. Then $N(z)N(w)=N(1)=1$, and both norms are nonnegative integers, forcing $N(z)=1$. Conversely, if $N(z)=1$, then $z\bar{z}=\pm 1$, so $z^{-1}=\pm \bar{z}$ belongs to the ring. Thus $z$ is a unit in a ring exactly when $N(z)=1$.

For negative $d$, the equation becomes $a^2+|d|b^2=1$. If $d=-1$, its integer solutions give the four units $\pm 1,\pm i$. If $d<-1$, the term $|d|b^2$ forces $b=0$, leaving $a=\pm 1$. HenceFor $d=2$, the element $u=1+\sqrt{2}$ has field norm $|1-2|=1$. Its powers $u^n$, $n\ge 0$, are all units by multiplicative function and are distinct, since their positive real values strictly increase with $n$. Thus $\mathbb{Z}[\sqrt{2}]$ has infinitely many units. This argument concerns the stated integer-coefficient ring; it does not replace it by a larger ring of algebraic integers.