Does $x_1^n + x_2^n + \dots + x_n^n = z^n$ have infinitely many primitive solutions in positive integers?

The OP wants to know if there are infinitely many primitive solutions in positive integers $x_i$ to,

$$x_1^n+x_2^n +\dots x_n^n = z^n$$

We have proof there are infinitely many such solutions for $n = (2,3,4,5)$, but not yet for $n=(7,8)$. Note: We'll skip $n=2$ or Pythagorean triples since that's well-known.

I. n = 3

Let $0<x_1<x_2<x_3$ and,

$$x_1^3+x_2^3+x_3^3 = x_4^3$$

then the complete positive integer solution was given by Choudhry's On Equal Sums of Cubes (1998). For the summarized version, given positive integers $(a,b,c)$, then

$$\begin{aligned} d\,x_1 &= (-a^3 - b^3 + c^3)c\\ d\,x_2 &= -(a^2 - a b + b^2)^2 + (a + b)c^3\\ d\,x_3 &= (a^2 - a b + b^2)^2 + (2a - b)c^3\\ d\,x_4 &= (a^3 + (a - b)^3 + c^3)c\end{aligned}$$

where $a>b$, and $c >(a^3+b^3)^{1/3}$, and $d$ is a rational scaling factor such that $\text{GCD}(x_1,x_2,x_3,x_4)=1$.

Note: See also this 2013 post, "Find all integer solutions to Diophantine equation $x^3+y^3+z^3=w^3$".

II. n = 4

Using an elliptic curve, there are infinitely many primitive solutions to

$$x_1^4+x_2^4+x_3^4+x_4^4 = (x_1\pm x_2\pm x_3\pm x_4)^4$$

Note: See also this 2024 post, "On why solutions to $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ also come in pairs".

III. n = 5

Given the 1934 parameterization by Sastry,

$$(x^5 + 25y^5)^5 + (x^5 - 25y^5)^5 + (10x^3y^2)^5 + (50x y^4)^5 + (-x^5 + 75y^5)^5 = (x^5 + 75y^5)^5$$

choose positive rationals $(x,y)$ such that,

$$y > \frac{x}{\sqrt[5]{75}}$$

where $\frac1{\sqrt[5]{75}} \approx 0.43$, then all terms are positive, and remove all common factors.

IV. n = 6

$$x_1^6+x_2^6+\dots+x_6^6 = z^6$$ $$x_1^6+x_2^6+\dots+x_\color{red}7^6 = z^6$$

No solution yet known for the former. For the latter, the largest primitive solution to used to be just $z \approx 0.4$ million found in 2010. This has been superseded by $z \approx 21.9$ million found in 2023, a value more than 50 times larger.

$$\small 19424433^6 +19128522^6 +13805652^6 +9771972^6 +8968470^6 +8306212^6 +5749170^6 = \color{red}{21934895^6}$$

How that was found is anybody's guess.

Note: See also this 2023 post, "On Sixth Powers $x_1^6+x_2^6+x_3^6+\dots+x_7^6 = z^6$ revisited".

V. n = 7,8

There are three primitive solutions for $n=7$ and one for $n=8$. There are probably infinitely many but, there is no proof, just hand-waving. However, there is an identity for

$$x_1^7+x_2^7+\dots+x_\color{red}8^7 = z^7$$

but some of the terms are not positive and its coefficients are roughly 3900-digit numbers.

Note: See also this 2018 post, "An elliptic curve $x(x - 78126^2)(x - 4\times5^7) = y^2$ for $x_1^7+x_2^7+\dots +x_8^7= z^7$"