On elliptic curves, $\sqrt{x^2-101y^2} ,\sqrt{x^2+101y^2}$, and their ilk 
I. Elliptic curves

Given integers $a,b,m_k$. Let,
$$x^2+a = m_1u_1^2\\x^2+b = m_1u_2^2\tag1$$
If there is a rational point $x_i$, then the pair (after a transformation) is birationally equivalent to an elliptic curve, call it $E_1$, and frequently has infinitely many rational points. Assume a second $m_2$,
$$x^2+a = m_2v_1^2\\x^2+b = m_2v_2^2\tag2$$
with a different rational point $x_j$, and yielding an elliptic curve $E_2$.

II. Question 

Q1: What is the relationship between $m_1,m_2$ given that ALL the rational points on $(1),(2)$ are on the single elliptic curve $E_3$,
$$(x^2+a)(x^2+b) = y^2\tag3$$
Or knowing the generators of $E_3$, can we predict what and how many square-free integer $m_k>1$ are permissible? 

III. Example

For simplicity, assume the special case of $b = -a$ which makes $m_1=1$ as the congruent number problem. Let $a=101$. For,
$$(x^2+a)(x^2-a) = y^2\tag4$$
two solutions are $x_1 = \frac{2015242462949760001961}{118171431852779451900}$, and $x_2 =\frac{2125141}{63050}$. As applied to,
$$x^2+101 = m_k u_1^2\\x^2-101 = m_k u_2^2\tag5$$
the point $x_1$ also solves $m_1 = 1$, while $x_2$ also solves $m_2 = 101$. (What other $m_k$ is permissible?)
Q2: In general, if $(5)$ is rationally solvable for one integer $m_k$, does it imply finitely (or infinitely) many other square-free integer $m_k > 1$? 
 A: The possible $m_k$, when assumed to be squarefree, must be divisors of the resultant of $x^2 + a$ and $x^2 + b$, which is $(a-b)^2$; so $m_k \mid a-b$. (Note that bot factors must be in the same square class; if a prime $p$ divides the squarefree representative of this class, then the binary forms $x^2 + a z^2$ and $x^2 + b z^2$ have a common root mod $p$.)
This is the usual argument used in 2-descent on elliptic curves or hyperelliptic Jaobians that one uses to show that the Selmer group
is contained in a finite subgroup of the multiplicative group of the
relevant etale algebra modulo squares (the algebra here is just
${\mathbb Q} \times {\mathbb Q}$; we are looking at a 2-isogeny).
Put differently, you get a homomorpism
$E_3(\mathbb Q) \to {\mathbb Q}^\times/{\mathbb Q}^{\times 2}$
(taking one of the points at infinity as the origin)
whose image gives you the $m_k$ for which there are rational points
on the covering curves. To figure out exactly which $m_k$ occur
may be difficult, since this is essentially equivalent to determining
the Mordell-Weil group of $E_3$ (for which no method is known so
far that could be shown to work in all cases).
