Find the unique cases when ${t}^{2} - 4$ is a perfect square say, ${n}^{2}$, with height bound $|t| \le N$ for positive integer $N \ge 1$, when $t$ is a rational where $t = p/q$ and integers $p$ an $q$ are relatively prime, $|p| \le N$ and $1 \le q \le N$. I am looking for a counting like solution of the unique cases, exact if possible or more likely an asymptotic expansion.


For a given $N$ there are $N \left({2\, N + 1}\right)$ cases to consider. The relatively prime condition reduces this by $6/{\pi}^{2}$ which is the probability that two random integer are relatively prime. I left off the next order term.
 For the integer case the only solutions are for $t = \pm 2$ which is covered in my other calculations.



Find the rational cases where ${t}^{2} - 4$ is a perfect square with height bound $|t| \le N$ for positive integer $N \ge 1$
From $t^2-4=s^2$ we get
$$
t^2-s^2=4~~ \Longrightarrow ~~ (t+s)(t-s) = 4
$$
hence the general rational solution $(t,s)$ is, putting $2\lambda = t+s$:
$$
\left( \lambda+\frac{1}{\lambda}, \lambda-\frac{1}{\lambda} \right).
$$
(It is easy to check that this indeed solves your equation.) So now we need to find the height of $t = \lambda + 1/\lambda$ in terms of the height of $\lambda=:n/d$. We have
$$
t = \lambda+\frac{1}{\lambda} = \frac{d^2+n^2}{dn},
$$
where the last fraction is obviously in lowest terms, at least if $n/d$ was. Hence the height of $t$ satisfies
$$
H(t) \sim H(\lambda)^2
$$ 
asymptotically. Since there are $O(N^2)$ rational numbers $\lambda$ of height $< N$, the number of $t$ with $H(t)<N$ that satisfies $t^2-4=\square$ is $O(N)$. 


