Are there general ways for given rational coefficients $a,b,c$ (I am particularly interested in $a=3,b=1,c=8076$, but in general case too) to answer whether this equation has a rational solution or not?

2$\begingroup$ The answer appears to be no. The given genus 1 curve has a Jacobian that is an elliptic curve with rank 1, but the given cover is in a Sha component. This can be shown by a 4descent, or maybe by a CasselsTate pairing. $\endgroup$ – MyNinthAccount Aug 10 '19 at 22:27
Your curve is a genus 1 curve, usually expressed as $$ y^2= abx^4+bc$$ (just by multiplying everything by $b$ and changing $y$ by $by$).
The curve has local points everywhere, but since it looks like has no rational points, you can try to do a 2descent. This can be done easily with the TwoCoverDescent algorithm as explained in
Nils Bruin and Michael Stoll. Twocover descent on hyperelliptic curves. Math. Comp., 78:23472370, 2009.
It is not guaranteed to succeed, but in the special case you are interested $a=3$, $b=1$ and $c=8076$ it does and the answer is that it has no rational points.
All this has been implemented in MAGMA. You can use the following code
P:=PolynomialRing(Rationals()); H:=HyperellipticCurve((3)*P.1^4+8076); Hk:=TwoCoverDescent(H);#Hk;
It answers 0, so there is no 2cover with rational points, so the original curve has no rational points.

$\begingroup$ Are you discussing the particular case $a=3$, $b=1$, $c=8076$ that the OP was particularly interested in? If so, you should say that's what you're doing. (I guess that's embedded in the MAGMA code, but for those of us who don't use Magma very often, it's not clear.) In general, as someone else listed as a comment, this curve represents an element of order 1 or 2 in the TateShafarevich group of its Jacobian. If order 1, then it has a rational point, if order 2, then not. $\endgroup$ – Joe Silverman Aug 10 '19 at 22:30

$\begingroup$ @JoeSilverman Thanks, I corrected to say I did the special case. Of course they can be interpreted as elements of the TateShafarevich group, but sometimes is not that easy to know if they are of order 1 or 2. There are other methods to do it, I just proposed one... $\endgroup$ – Xarles Aug 10 '19 at 22:37
Multiplying by $x^2$ and denoting $X:=abx^2$, $Y:=ab^2xy$, we get an elliptic curve: $$Y^2 = X^3ab^2cX.$$ If it turns out that this curve has zero rank, then the number of solutions is finite and they are given by torsion points.
Luckily this is the case for the given values $a,b,c$, while the torsion points do not give a solution to the original equation.

$\begingroup$ "the torsion points do not give a solution to the original equation"  is it somehow clear? $\endgroup$ – Fedor Petrov Aug 10 '19 at 22:40

3$\begingroup$ The 2torsion point (only one that exists here) has $(X,Y)=(0,0)$, and it reverses to $x=0$, not a solution.. On the other hand, the elliptic curve in question has rank 1 (generated by $X=110^2/13^2$), not rank 0. The problem then reduces to whether any relevant $X$values in the elliptic curve solutions are square (reversing to a rational $x$value), which is likely as hard as (or equivalent to) the original problem. $\endgroup$ – MyNinthAccount Aug 10 '19 at 23:41

$\begingroup$ @MyNinthAccount: Indeed. I've mistyped the numbers while checking the rank. $\endgroup$ – Max Alekseyev Aug 11 '19 at 9:00
Above equation shown below:
$ax^4+by^2=c$ $(1)$
For equation $(1)$, Seiji Tomita has given parametric solution.
For given $(a,b,c) =[(2),(1),(3*33^4)]$ the numerical solution,
for equation (1) is shown below:
$(x,y)=(13,1871)$
For further details see his web site & the link is given below:
http://www.maroon.dti.ne.jp/fermat
Click on computational number theory & then select article #327

2$\begingroup$ And how does this answer the question? $\endgroup$ – András Bátkai Aug 26 '19 at 17:25