Suppose that $\alpha \in \mathbb{Z}[\sqrt{7}]$ where $\alpha$ is of the form $a + b\sqrt{7}$ where $a, b \in \mathbb{Z}$. Because $\alpha$ is a unit if and only if $N(\alpha)=\pm 1$ we must show:
$N(\alpha) = a^2 - 7b^2 = \pm 1$
has infinitely many solutions for $\alpha$. Not quite sure how show this. I am trying to find an $\alpha$ that satisfies the equation and then use the fact that $\mu^n$ is a unit if $\mu$ is a unit.
On
Remember that $(-1)^n$ is $1$ or $-1$ depending on whether $n$ is odd or even. If you can find just one unit other than $1$ or $-1$ with norm $\pm1$, then you can find infinitely many other units by simply raising that unit to integer powers.
In the case of $\mathbb{Z}[\sqrt{7}]$ you should be able to quickly find $8 + 3 \sqrt{7}$ just by trial and error. (In cases like $\mathbb{Z}[\sqrt{19}]$, it would be better to use Pell's equation, as trial and error would take annoyingly long).
Now notice that $(8 + 3 \sqrt{7})^2 = 127 + 48 \sqrt{7}$. This has a norm of $1$. What is $(8 + 3 \sqrt{7})^3$? $(8 + 3 \sqrt{7})^4$? You get the idea.
And of course $(8 + 3 \sqrt{7})^0 = 1$. Try negative exponents, too.
What about $(8+3\sqrt 7)^n$, $n\in\mathbb Z$? (For finding all units the key words are "Pell's equation".)