Let us fix a rational number $c=\frac{a}{b}$ where (a) the square root of $c$ is irrational, and (b) $a$ and $b$ are both positive integers no larger than say, 10. [The integer $b$ may be 1.]
Does the diophantine equation
$$n^2 = cm^2 + 1 $$
have an infinite number of solutions i.e., an infinite number of pairs $(m,n)$ so that the above equation holds--for any such choice for $c$? I'd be really surprised if there were such a $c$ that yielded an infinite number of solutions, but I cannot prove or disprove either way.
If this is too broad a question, I would be quite happy if one could prove or disprove whether
$$n^2 = \frac{4m^2}{3} + 1 $$
has a bounded number of solutions.
There’d be an infinite number of (integer) solutions once there is a single non-trivial (i.e., nonzero) solution, say $(m_0,n_0)$. Note that necessarily, in this case we must have $b|m_0^2$ (assuming $a$ and $b$ are coprime). To see this observe that you have $$1=(n_0^2-cm_0^2)^2=(n_0^2+cm_0^2)^2-c(2m_0n_0)^2\,,$$ which gives another (larger, magnitude-wise) solution $$m_1:=2m_0n_0\,,\,~\,~\,~ n_1:= n_0^2+cm_0^2 \,.$$ Note that $n_1$ is an integer.
(Edit: It turns out that a nontrivial solution can be found by first simply solving the Pell equation $$n^2-abm^2=1\,.$$ Note that $ab$ cannot be a square. Granted a nontrivial solution to this, say $ (m’,n’)$, then a nontrivial solution to your problem becomes $(m_0,n_0):=(bm’,n’)$.))
For instance, for the example you gave—$3n^2=4m^2+3$—we have the non-trivial solution $$(m_0,n_0)=(6,7)\,,$$ which quickly leads to the infinitude of solutions via the above construction $$(6,7)\,,(97,84)\,,(18817,16286)\,,\cdots\,.$$