The Diophantine equation,

$$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k\tag1$$

for either $k=5$ or $6$ is quite well explored, and it has long been known that it has an infinite number of primitive solutions (as points on an elliptic curve, or an infinite family of polynomials,). However,

$$x_1^k+x_2^k+x_3^k+x_4^k = y_1^k+y_2^k+y_3^k+y_4^k\tag2$$

for either $k=7$ or $8$, results seem to be harder to find. Only one solution found back in 2006 is known for $k=8$. Many solutions exist when $k=7$, but it is not known if there are infinite many. Choudhry reduced $(2)$ when it is true for $k=1,3,7$ plus a fourth constraint (re Wolfgang's comment below), namely $\sum\limits^4 x_i = 0$ (and since $k=1$, then also $\sum\limits^4 y_i=0$ ), to a multi-variable cubic equation by passing through the simultaneous equations, in Choudhry's notation,

$$\begin{aligned} X_1X_2X_3\,&=Y_1Y_2Y_3\\ 2(X_1^4 + X_2^4 + X_3^4) - 5(X_1^2 + X_2^2 + X_3^2)^2\, &= 2(Y_1^4 + Y_2^4 + Y_3^4) - 5(Y_1^2 + Y_2^2 + Y_3^2)^2 \end{aligned}\tag3$$

Using that cubic, Choudhry (3) and Wroblewski (21) found a total of $3+21=24$ solutions to $(2)$ valid for $k=1,3,7$. Results can be found in euler.free.fr. Using the $x_i,y_i$, I derived their $X_i, Y_i$. Surprisingly, seven had a simple linear constraint, namely $X_3=2Y_3$. Let,

$$X_1,\,X_2,\,X_3 =u_1 u_2,\;u_3 u_4,\;2t$$ $$Y_1,\,Y_2,\,Y_3 =2u_1 u_4,\;u_2 u_3,\;t$$

then solutions are,

$$\begin{array}{|c|c|c|c|c|c|} \text{#}&u_1&u_2&u_3&u_4&t\\ 1&227 &27 &113 &13 &1305\\ 2&431 &187 &365 &49 &6929\\ 3&127 &139 &313 &1 &23647\\ 4&303 &304 &338 &37 &24616\\ 5&871 &163 &4364 &127 &254405 \\ 6&439 &2459 &247 &1175 &261851 \\ 7&14737 &139 &15899 &25 &544069 \\ \end{array}$$

It was hard to find commonalities for the other seventeen solutions.

Questions:

1. Can the simultaneous eqns $(3)$, or the cubic described by Choudhry in the paper, be reduced to an elliptic curve?
2. Why does almost $1/3$ of the $24$ known solutions have $X_3=2Y_3$? Translated into the addends $x_i,y_i$ of $(2)$, this is equivalent to the 5th constraint that $x_1+x_2 = 2y_1+2y_2$. Is there some identity behind it? For example, I found this 7th deg multi-grade,$$1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = \\(-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k$$ for $k = 1,3,5,7,\;$ if $x^2-10y^2 = 9$, guaranteeing an infinite supply with the constraint $x_2 = 5x_1$, though I doubt anything for $(2)$ would be as simple as this.

Note:

However, the fifth degree version,

$$x_1^5+x_2^5+x_3^5 = y_1^5+y_2^5+y_3^5$$

$$x_1+x_2+x_3 = y_1+y_2+y_3=0$$

does have a polynomial identity behind it, found by Choudhry and Wrobleski. (See "A quintic Diophantine equation with applications to two Diophantine systems concerning fifth powers")

P.S. Eq.1 is briefly discussed in this MO post.