I have been studying the equation $a^6+b^6+c^6=d^2$, trying to find rational solutions. I know it is a K3 surface, with high Picard rank, so there should be rational or elliptic curves on it.

When Elkies found solutions to the equation $a^4+b^4+c^4=d^4$, he started by using the simpler equation $r^4+s^4+t^2=1$. Taking inspiration from this, I looked at the equation $(2):y^2=x^3+z^6+1$. Noting the two trivial solutions $(x,y)=(-1,z^3)$ and $(x,y)=(-z^2,1)$, and taking the sum of the two points (in the elliptic curve addition sense), and multiplying to remove fractions yields the parametric equation:

$(3z^6+9z^5+15z^4+17z^3+15z^2+9z+3)^2=(2z^4+4z^3+5z^2+4z+2)^3+(z^2+z)^6+(z+1)^6$

This would result in a solution to my original problem, if $2z^4+4z^3+5z^2+4z+2$ were a square. This yields an equation $u^2=2z^4+4z^3+5z^2+4z+2$. By inspection I found the solution $(z,u)=(-1,\pm1)$. Unfortunately, if $z=-1$, then the equation above collapses into $1^2=1^6+0^6+0^6$, which is trivial. However, this equation can be converted into Weierstrass form, resulting in: $y^2=x^3-x^2-8x-4$, with the point $(-1,1)$ taken to be the point at infinity, and the other point $(-1,-1)$ taken to the point $(-2,0)$.

However, this elliptic curve has only those two rational points on it, therefore no solutions to the original equation can be obtained. Other points I have found on equation (2) by using chord and tangent methods starting from those two initial points result in parametric equations with a polynomial of too high order to result in an elliptic curve.

What are some other more fruitful approaches to this problem? Note that I am aware of other questions on mathoverflow providing some solutions. However, I am looking for a way to generate infinitely many solutions. This would preferably be with a parametric equation, however I'll also be happy with an elliptic curve and a rational point of infinite order.

If at all possible I would appreciate hints in the right direction over full solutions. I'm wanting to use this to grow my expertise and problem solving ability in this area. I'll update this question with any future attempts that are worth mentioning.