$a,b \in \mathbb Z$ we have that $a+ b\phi \in \mathbb Z[\phi]^* \iff a^2 +ab-b^2 = \pm 1$

For $z = a + b \phi$, we write $\overline{z} = a + b \overline{\phi}$.

Define the function $N: \mathbb{Z}[\phi] \rightarrow \mathbb{Z}$ by $N(z)=z \overline{z}$. It's easy to check that $N(a+b \phi) = a^2 +ab -b^2 \in \mathbb{Z}.$

Proposition: $z \in \mathbb{Z}[\phi]$ is a unit if, and only if, $N(z) \in \mathbb{Z}$ is a unit.

Proof:

($\Rightarrow$) Suppose $(a+b\phi)(c+d\phi)=1$. By the hint we have $(a+b\overline{\phi})(c+d\overline{\phi})=1$, and so $$N(a+b\phi)N(c+d\phi) = (a+b\phi)(a+b\overline{\phi})(c+d\phi)(c+d\overline{\phi}) = 1.$$ Since $N(a+b\phi)$, $N(c+d\phi) \in \mathbb{Z}$, we must have either $$N(a+b\phi) = N(c+d\phi) = 1 \ \ \text{or} \ \ N(a+b\phi) = N(c+d\phi) = -1.$$

($\Leftarrow$) Suppose that $N(a+b\phi) = (a+b\phi)(a+b\overline{\phi}) = (-1)^k, k\in \{ 0,1 \}$.

We want to find $c,d \in \mathbb{Z}$ such that $c+d\phi = a+b\overline{\phi}$. So write $$ c+d\phi = c+ \frac{1}{2} d + \frac{\sqrt{5}}{2} \equiv a + \frac{1}{2}b -\frac{\sqrt{5}}{2} = a+b\overline{\phi},$$ and equate coefficients in $\mathbb{Z} + \sqrt{5}\mathbb{Z}$ to get $c = a+b$ and $d=-b$. That is, we have $$(a+b\phi)((-1)^k (a+b) + (-1)^{k+1} b\phi) = 1.$$

$\hspace{18cm} \square$

Therefore, given $a+b\phi \in \mathbb{Z}[\phi]$, $$a+b\phi \in \mathbb{Z}[\phi]^* \Leftrightarrow a^2 +ab -b^2 = N(a+b\phi ) = \pm 1.$$