Finding multigrade $(8,4,4)$ solutions satisfying $a^8+b^8+c^8+d^8=e^8+f^8+g^8+h^8$

A while ago, @Aleksandr posed this question, about finding new solutions to $$a^8+b^8+c^8+d^8=e^8+f^8+g^8+h^8$$ (where the solutions should be non-trivial and primitive) and he stated the known result that in 2006, Nuutti Kuosa discovered $$1953^8+2012^8+3113^8+861^8=1128^8+2767^8+2557^8+2823^8.$$ I investigated with a computer search and found no additional solution with a sum less than $11677^8$. So the solution above remains to be the only known one.

My question regards multigrade solutions to the equation, i.e. solutions to $$a^k+b^k+c^k+d^k=e^k+f^k+g^k+h^k \qquad \textrm{for } k=2,4,8\tag{1}$$ This question is inspired by @TitoPiezasIII, who asked this question, giving parameterizations for a subset of near-misses. In the subset, he required the additional condition of $$a^2+b^2=e^2+f^2 \tag{2}.$$

It turns out, that the system of equations of $(1),(2)$ is entirely equivalent (when asking for non-trivial solutions) to the system below: \begin{align} a^4+a^2b^2+b^4 &= g^4+g^2h^2+h^4 \tag{3} \\ c^4+c^2d^2+d^4 &= e^4+e^2f^2+f^4 \tag{4} \\ a^2+b^2&=e^2+f^2 \tag{5} \\ c^2+d^2&=g^2+h^2 \tag{6} \end{align} Giving rise to equations that have been parameterized (partly) in posts like this and this. Two of the parameterizations for equations $(3)$ (and also $(4)$ of course) are \begin{equation} \begin{array}{cc} \textbf{Cubic parametrization} & \textbf{Quartic parametrization}\\ \begin{aligned} a&=2m^3+m^2+3m-1,\\ b&=-m^3+3m^2+m+2,\\ g&=-2m^3+m^2-3m-1,\\ h&=m^3+3m^2-m+2, \end{aligned} & \begin{aligned} a&=m^4-3m^3+m^2-m+2,\\ b&=-2m^4-m^3-m^2-3m-1,\\ g&=m^4+3m^3+m^2+m+2,\\ h&=-2m^4+m^3-m^2+3m-1. \end{aligned} \end{array} \end{equation}

My question is this: Can this alternative system of equations yield multigrade $(8,4,4)$ solutions.

I guess if I had to make the question less open, it would be: If one parameterizes equation $(3)$ by the cubic/quartic above and equation $(4)$ by a different cubic/quartic above, can we find parameters such that equations $(5)$ and $(6)$ also hold?

Any results are interesting, showing that for example using the cubic parameterizations for both $(3)$ and $(4)$ does not give rise to any solutions rules out a lot of the (known) possibilities for any solutions.

Also, if you want to chat about this, the chatroom Equal sums of powers (view transcript, join the room) is open for all these types of diophantine equations, with a leaderboard here.