Solvability of $a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$ when $n \geq 5$ is prime? Aware of a Darmon-Merel theorem that asserts that if $n \geq 5$ is prime then the equation $a_{1}^{2} = a_{2}^{n} + a_{3}^{n}$ has no solution in relatively prime integers $a_{1}, a_{2}, a_{3},$ I wonder if this is also true, under the same conditions, for the equation
$$a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}?$$
If there is already any known result in existence, bringing it in is more than welcome.

For $0<a\leq b\leq c\leq 100$ with $\gcd(a,b,c)=1$, we have the sums including the one by user R. Israel,
$$3^5+ \color{brown}{49}^5+ 69^5 = 42971^2$$
$$20^5+ 68^5+  \color{brown}{81}^5 = 70313^2$$
$$19^5+ 80^5+ 97^5 = 108934^2$$
$$ \color{brown}{16}^5+ 83^5+ 97^5 = 111926^2$$
While your question was on prime $n\geq 5$, there is also for $n=6$,
$$42^6+  \color{brown}{81}^6+  \color{brown}{100}^6 = 1134865^2$$
(It's quite interesting that a lot of the addends are squares.)
For $n=7,8$, there are none in that range (anyone can extend it?), though there is the near-miss,
$$(a^2-b^2)^8+(a^2+b^2)^8+(2ab)^8 = 2(a^8+14a^4b^4+b^8)^2$$
