There are infinitely many. Let $V$ be the subspace of $H$ where the trace is zero. Then norm gives an nondegenerate quadratic form on $V$. For any $v \in V$, the field $\mathbb{Q}(v)$ is isomorphic to $\mathbb{Q}(\sqrt{-N(v)})$. Recall that the fields $\mathbb{Q}(\sqrt{D_1})$ and $\mathbb{Q}(\sqrt{D_2})$ are isomorphic if and only if $D_1/D_2$ is a square.

So we need to show that $V$ takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$. For example, if we are dealing with the standard quaternion algebra, we need to show that $p^2+q^2+r^2$ takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$. This is easy enough that probably any method you think of will work. Here is what I came up with: Take $u$ and $v$ linearly independent members of $V$. Let $a=u^{-1} v$.

First, suppose that $-N(a)$ is a square, say $k^2$. Then, for $s$ and $t \in \mathbb{Q}$, we have $N(su+tv) = N(u) N(s+t a v)= N(u) (s-kt)(s+kt)$, and this expression clearly takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$ as we vary $s$ and $t$.

If $N(a)$ is not a square, let $K=\mathbb{Q}(a)$, this is a subfield of $H$ isomorphic to $\mathbb{Q}[\sqrt{N(a)}]$. For $b \in K$, we have $N(bu) = N_{K/\mathbb{Q}}(b) N(u)$, and $bu$ is in $V$. Since there are infinitely many primes that split principally in $K$, there are infinitely primes that occur as norms $ N_{K/\mathbb{Q}}(b)$, and thus we get an infinite subgroup of $\mathbb{Q}^*/(\mathbb{Q}^*)^2$.

There are infinitely many. Let $V$ be the subspace of $H$ where the trace is zero. Then norm gives an nondegenerate quadratic form on $V$. For any $v \in V$, the field $\mathbb{Q}(v)$ is isomorphic to $\mathbb{Q}(\sqrt{-N(v)})$. Recall that the fields $\mathbb{Q}(\sqrt{D_1})$ and $\mathbb{Q}(\sqrt{D_2})$ are isomorphic if and only if $D_1/D_2$ is a square.

So we need to show that $V$ takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$. For example, if we are dealing with the standard quaternion algebra, we need to show that $p^2+q^2+r^2$ takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$. This is easy enough that probably any method you think of will work. Here is what I came up with: Take $u$ and $v$ linearly independent members of $V$. Let $a=u^{-1} v$.

First, suppose that $-N(a)$ is a square, say $k^2$. Then, for $s$ and $t \in \mathbb{Q}$, we have $N(su+tv) = N(u) N(s+t a)= N(u) (s-kt)(s+kt)$, and this expression clearly takes infinitely many values in $\mathbb{Q}^*/(\mathbb{Q}^*)^2$ as we vary $s$ and $t$.

If $N(a)$ is not a square, let $K=\mathbb{Q}(a)$, this is a subfield of $H$ isomorphic to $\mathbb{Q}[\sqrt{N(a)}]$. For $b \in K$, we have $N(ub) = N_{K/\mathbb{Q}}(b) N(u)$, and $ub$ is in $V$. Since there are infinitely many primes that split principally in $K$, there are infinitely primes that occur as norms $ N_{K/\mathbb{Q}}(b)$, and thus we get an infinite subgroup of $\mathbb{Q}^*/(\mathbb{Q}^*)^2$.