infinitely many units in $\mathbb{Z}[\sqrt{d}]$ for any $d>1$.

I don't actually know any of the standard proofs of Dirichlet's unit theorem that Qiaochu Yuan refers to, but I think they might use Minkowski's theorem about the existence of nonzero lattice points in sufficiently large centrally symmetric convex subsets of $\mathbf R^n$. I'll give a proof for the specific question asked here, which uses Minkowski's theorem, but only in the very simple case of a parallelogram. I'll write the proof top-down, so that one sees how the theorem is used before I'll state (and prove) it.

First I recall some generalities about the ring $R=\mathbf Z[\sqrt d]$ for $d$ a positive squarefree number, to get started. The additive group of $R$ is free Abelian of rank $2$ with generators $1,\sqrt d$. The norm map $N:R\to\mathbf Z$ given by $a+b\sqrt d\mapsto a^2-db^2$ is multiplicative, and the units of $R$ are precisely the elements with norm $\pm1$. One has the non-trivial unit $-1$, but it is of finite order; the point to prove is therefore that the subgroup of positive units is non-trivial (once a positive unit${}\neq1$ is found, its powers form an infinite set of units). I will reason by contradiction, so assume that $1$ is the unique positive unit of $R$. The first step will be to show that this would imply that for any $n\in\mathbf N$, the number of positive $r\in R$ with $|N(r)|=n$ is finite, in fact at most $n^2$.

Lemma. For any $n\in\mathbf N_{>0}$, the number of principal ideals of $R$ that contain $n$ is at most $n^2$.

Proof. Since these ideals all contain $nR$, they all map to principal ideals of $R/nR$, and the mapping is injective. The number of principal ideals of $R/nR$ cannot exceed the number $n^2$ of its elements. QED

The bound given here is far from sharp, but finiteness is all we need. Under the hypothesis that $1$ is the unique positive unit of $R$, two positive elements of $R$ generate distinct principal ideals, and if $|N(r)|=n$, the ideal generated by $r$ contains $n$, so the lemma justifies our claim that the number of of such$~r$ is finite.

This means that for every $M>0$ there is some $\varepsilon_M>0$ such that for all $a,b\in\mathbf Z$ with $0\neq|a+b\sqrt d|<\varepsilon_M$ one has $|N(a+b\sqrt d)|\geq M$. We shall show that for sufficiently large $M$ this contradicts Minkowski's theorem. Note that $N(a+b\sqrt d)=(a+b\sqrt d)(a-b\sqrt d)$, so we can bound $N(a+b\sqrt d)$ above if in addition to the value $a+b\sqrt d$ we also bound its conjugate $a-b\sqrt d$. Now the linear endomorphism of $\mathbf R^2$ sending $\binom ab\mapsto\binom{a+b\sqrt d}{a-b\sqrt d}$ has determinant $-2\sqrt d$, so the conditions $|a+b\sqrt d|<x$ and $|a-b\sqrt d|<y$ define the interior of a parallelogram of area $4\frac{xy}{2\sqrt d}$, for any $x,y>0$. Minkowski's theorem says such a parallelogram will contain a nonzero lattice point whenever its area is greater than $2^2=4$.

So here is how to obtain a contradiction: take any $M>2\sqrt d$ and put $$ x_0=\min\{ z\in R \mid z>0 \land |N(z)|\leq M\}\qquad\text{and}\qquad y_0=\frac M{x_0}. $$ Then $\frac{x_0y_0}{2\sqrt d}>1$, so Minkowski's theorem ensures the existence of $a,b\in\mathbf Z$, not both $0$, with $|a+b\sqrt d|<x_0$ and $|a-b\sqrt d|<y_0$. But then on one hand $|N(a+b\sqrt d)|>M$ by the choice of $x_0$ (and the fact $N(-z)=N(z)$), but on the other hand $|N(a+b\sqrt d)|=|a+b\sqrt d||a-b\sqrt d|<x_0y_0=M$, a contradiction.

Minkowski's theorem. Any centrally symmetric convex subset $S$ of $\mathbf R^d$ of volume greater than $2^d$ contains a nonzero element of $\mathbf Z^d$.

Proof. The map $f:\mathbf R^d\to(2\mathbf Z)^d$ is locally area-preserving, so its restriction to $S$ cannot be injective since the total area at arrival is $2^d$. If $s,s'\in S$ have $s\neq s'$ and $f(s)=f(s')$ then by central symmetry $-s'\in S$, and by convexity $\frac{s-s'}2\in S$; therefore, since $f(s-s')\in(2\mathbf Z)^d$, one has $\frac{s-s'}2\in S\cap(\mathbf Z^d\setminus\{0\})$. QED