As a natural extension of the question titled Solvability of $a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$ when $n \geq 5$ is prime?, I wonder if the equation $$2a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$$ is solvable for co-prime $a_i$ when $n \geq 5$ is prime?
The Darmon-Merel theorem asserts that $a_{1}^{2} \neq a_{2}^{n} + a_{3}^{n}$ for all $n \geq 3$ and all integers $a_{1}, a_{2}, a_{3}$ not all $0$, but I am not sure about the result when $2a_{1}^{2}$ is considered instead.
If there is already any known result in existence, bringing it in is more than welcome.
I do not know why co-prime solutions are more interesting and must these be mutual co-prime or co-prime altogether, but there are few mutual co-prime ones $a_1,a_2,a_3<1000$ (not sure all of these found due to precision trunks):
$$\begin{align}4^5+239^5+659^5=2\cdot 7907819^2\\ 40^5+ 617^5+ 633^5= 2\cdot 9773625^2\\ 73^5+ 121^5+ 144^5= 2\cdot 212047^2\\ 96^5+ 709^5+ 877^5= 2\cdot 18681029^2\\ 161^5+ 380^5+ 479^5= 2\cdot 4077240^2\\ 211^5+335^5+704^5=2\cdot 9422435^2\\ 224^5+387^5+541^5=2\cdot 5271996^2\\ 360^5+493^5+517^5=2\cdot 6004405^2\\ 369^5+640^5+881^5=2\cdot 17957625^2\\ \end{align}$$