For $d$ not congruent to $3$ modulo $4$, the elements of $\mathcal{O}_K$ are of the form $a+b\sqrt{-d}$ with $a,b\in\mathbb{Z}$; the map $a+b\sqrt{-d}\mapsto (a+b\sqrt{-d})(a-b\sqrt{-d}) = a^2+db^2$ is multiplicative and has positive integers as images, hence $a+b\sqrt{-d}$ is a unit if and only if its image is $1$. But $a^2+db^2=1$ with $d\gt 1$ requires $b=0$ and $a=\pm 1$. If $d=1$, then we also get $a=0$, $b=\pm 1$, i.e., $i$ and $-i$.
For $d\equiv 3 \pmod{4}$, the element sof $\mathcal{O}_k$ are of the form $\frac{a+b\sqrt{-d}}{2}$ with $a,b\in\mathbb{Z}$ of the same parity. The map $\frac{a+b\sqrt{-d}}{2} \longmapsto \left(\frac{a+b\sqrt{-d}}{2}\right)\left(\frac{a-b\sqrt{-d}}{2}\right) = \frac{a^2+db^2}{4}$ is multiplicative and has positive integers as their images, hence an element of $\mathcal{O}_K$ is a unit if and only if $a$ and $b$ are of the same parity and $a^2+db^2=4$. If $d\gt 3$ then $d\gt 7$, so this requires $b=0$ and $a=\pm 2$, hence the element in question is $\pm 1$.
If $d=3$, then we also have the solution $a=\pm 1$, $b=\pm 1$, which gives $\pm\frac{1+\sqrt{-3}}{2},\quad \pm\frac{1-\sqrt{-3}}{2}.$