Why $4b^3p=2a^2+qb$ has no solutions for integers $a,b$ and two odd primes $p<q$?

Question

Empirically I do not find any integer solutions for $4b^3p=2a^2+qb$, where $a$ and $b$ are non-zero integers and $2<p<q$ are two odd primes.

If it is provable, why this equation does not provide a solution?

Answer 1

Non-trivial solution: $p=3,q=47,a=1,b=2$

Answer 2

If $b=0$, we quickly find $a=0$, which indeed gives a solution.

Write $b=uv^2$ where $u$ is square-free. Then from $$2a^2=uv^2\cdot{(4u^2v^4p-q)} $$ we conclude that $v\mid a$ and write $a=vw$: $$2w^2=u\cdot\underbrace{(4u^2v^4p-q)}_{\text{odd}} $$ We conclude $u=2t$ and so $$w^2=t\cdot{(16t^2v^4p-q)} $$ It seems promising to find solutions with $w=1$ and $t=\pm1$, which amounts to looking for primes $p$ and integers $v$ such that one of $(2v)^4p\pm1$ is prime. Even with these additional restrictions we find lots of solutions $(p,q,a,b)$, such as $$ (3,47,1,2)^1, (5,79,1,2), (7, 113, 1, -2), \ldots, (41, 409999, 5, 50),\ldots,(97, 1553, 1, -2),\ldots, (983, 15727, 1, 2), (991, 64946177, 8, -128), (997, 1292113, 3, -18),\ldots (1000003, 38416115249, 7, -98),\ldots, (1000000007, 256000001791, 2, 8),\ldots$$

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$^1$ see answer by cr001.