Show by an elliptic curve $y^3=x^3-pqx$ that the product of two odd primes $p<q$ is a congruent number

Question

I am investigating whether a natural number $n=p\cdot q$ being a product of two distinct odd primes $p<q$ is a congruent number. For this I am re-formulating the question: Does an elliptic curve of the form $y^2=x^3-pqx$ has at least a rational points if $p<q$ are two distinct odd primes?

What I did so far:

Let $L_{a,b}:y=\frac{a}{b}x$ be a linear function where $a\in \mathbb{Z},b\in \mathbb{N}$ and $(a,b)=1$. To find rational points on the elliptic curve $E_{p,q}:y^2=x^3-pqx$ for fixed $(p,q)$, it is sufficient to solve for $(a,b)$. Conversly, if we fix a pair $(a,b)$, we may want to describe all pairs $(p,q)$ whose corresponding curve $E_{p,q}$ intersects $L_{a,b}$. To proceed in an elementary way, this is equivalent to solve the equation

$$x^3-\left(\frac{a}{b}\right)^2x^2-pqx=0$$

This cubic has $3$ solutions, one trivial given by $x=0$ and the others given by

$$ x=\frac{1}{2}\left(\frac{a}{b}\right)^2\pm\sqrt{\frac{\left(\frac{a}{b}\right)^4+4pq}{4}} $$

In order for $x$ to be rational, we ask that $a^4+4pqb^4$ is a square, say $$a^4+4pqb^4=c^2$$ for some integer $c$. This last equation can be factored as $4pqb^4=c^2-a^4=(c-a^2)(c+a^2)$. We are now left with the study of several cases given by the number of ways to assign the divisors of $4pqb^4$ to each of the factors $(c\pm a^2)$. Counting the number of cases boils down to computing half of the number of divisors of $2^2pqb^4$. For $\tau(n)$ the divisor function and $n=p_1^{e_1}\cdots p_k^{e_k}$, we have the following identity $\tau(n)=(e_1+1)\cdots(e_k+1)$. In our case we get at least

$$\frac{1}{2}\tau(2^2pqb^4)=\frac{1}{2}(2+1)(1+1)(1+1)(4+1)=30\, \text{cases}$$

since $b$ is not necessarily prime. The corresponding cases are:

\begin{array}{rlrlrlrl} 1&(22pqbbbb,\emptyset)&9&(22qbbbb,p)&16&(22pbbbb,q)&23&(2b,2pqbbb)\\ 2&(2pqbbbb,2)&10&(2qbbbb,p2)&17&(2pbbbb,q2)&24&(2bb,2pqbb)\\ 3&(pqbbbb,22)&11&(qbbbb,p22)&18&(pbbbb,q22)&25&(2bbb,2pqb)\\ 4&(bbbb,22pq)&12&(qbbb,p22b)&19&(pbbb,q22b)&26&(2bbbb,2pq)\\ 5&(bbb,22pqb)&13&(qbb,p22bb)&20&(pbb,q22bb)&27&(22b,pqbbb)\\ 6&(bb,22pqbb)&14&(qb,p22bbb)&21&(pb,q22bbb)&28&(22bb,pqbb)\\ 7&(b,22pqbbb)&15&(q,p22bbbb)&22&(p,q22bbbb)&29&(22bbb,pqb)\\ 8&(\emptyset,22pqbbbb)&&&&&30&(22bbbb,pq) \end{array}

From this emerge the following six conditions (see this Table for all $30$ undistilled, partially overlapping cases):

1. $\boldsymbol{q=a^2+pb^4}$: For $b=1$ and $a=2$ all consecutive primes $p,q$ give a solution whose difference is $4$. According to Polignac's conjecture (which isn't proved), there are infinitely many examples per any even difference of both primes.
2. $\boldsymbol{pq=a^2+b^4}$: This boils down to verifying the conjecture that $pq=a^2+b^4$ has infinitely many solutions, which brings us to a theorem of Iwaniec and Frielander (they state that there are infinitely many primes of the form $p=a^2+b^4$).
3. $\boldsymbol{pq=\frac{a^2+1}{4b^4}}$: The smallest found example for $b>1$ is $y^2=x^3-3281x$, which means $p=17,q=193$ and $a=1432,b=5,c=2050626$ and the discriminant $\Delta=\frac{1051266747969}{625}$. Two rational points on this curve are $(82025,23491960)$ and $\left(-\frac{1}{25},-\frac{1432}{125}\right)$
4. $\boldsymbol{p=qb^4-a^2}$: We note that if $b$ is odd then $a$ must be even. If $b$ is even, then $2^4|p+a^2$. This case can be transformed to the problem describing primes of the form $x^2+ny^2$ which is extensively elaborated by David A. Cox.
5. $\boldsymbol{p=\frac{2a^2+qb}{4b^3}}$: Some examples are $p=5,q=31,a=7,b=2$ and $p=7,q=31,a=9,b=2$ and $p=5,q=71,a=3,b=2$.
6. $\boldsymbol{pq=b^4-a^2}$: One example is $p=3,q=15$ where $a=1,b=2,c=31$ and $\Delta=\frac{961}{64}$. More generally, we need $p|b^2-a$ or $p|b^2+a$ and the same conditions apply to $q$.

My detailed question:

Is it possible to combine these six above-shown conditions to a "more relaxed" conditions which facilitates showing that the product $p\cdot q=n$ of two odd primes $p<q$ is a congruent number?

Precisely: Is it true that no product of two odd primes $p<q$ exist for which all these six above-shown contitions fail? Inversly stated, is it true that for two odd primes $p<q$ at least one of the above-shown conditions matches.

Or are we here on a pathway to something that isn't doable at all?