Does $x_1^8+x_2^8+x_3^8+x_4^4 = y_1^8+y_2^8+y_3^8+y_4^4$ have infinitely many solutions?

I. Question

Consider the following independent and symmetric mixed equations,

1. $\quad x_1^8+x_2^8+x_3^4 = y_1^8+y_2^8+y_3^4$
2. $\quad x_1^8+x_2^8+x_3^4+x_4^4 = y_1^8+y_2^8+y_3^4+y_4^4$
3. $\quad x_1^8+x_2^8+x_3^8+x_4^4 = y_1^8+y_2^8+y_3^8+y_4^4$
4. $\quad x_1^8+x_2^8+x_3^8+x_4^8+x_5^4 = y_1^8+y_2^8+y_3^8+x_4^8+y_5^4$

Since (without scaling) there is only one known integer that is the sum of four $8$th powers in two ways, or a $(8,4,4)$, it seems surprising there are infinitely many primitive solutions to the first, second, and fourth equations. Note: By primitive, we mean $\text{GCD}(x_i)=1$.

Q: Is it also true for the third equation?

II. Parameterizations

Many solutions to the third equation were kindly provided by Aleksandr and Joris. It was hard to find a pattern in all the numerical data, but I finally noticed the form $a^2+b^2 = c^2+d^2$. So we have the multigrade,

$$a^k + b^k + (u + v)^{k/2} + (u - v)^{k/2} = c^k + d^k + (u + w)^{k/2} + (u - w)^{k/2}$$

valid for $k=(2,4,8)$ if we find five integers $(a,b,c,d,w)$,

$$a^2+b^2 = c^2+d^2$$

$$u = \frac{\sqrt{a^4+a^2b^2+b^4-w^2}}{\sqrt3}$$

$$v = \frac{\sqrt{-a^4-b^4+c^4+d^4+2w^2}}{\sqrt2}$$

such that $(u,v)$ are also integers. This provides solutions to the second equation. But to fulfill the third, we require that one sign of $u \pm v$ and $u \pm w$ yield two squares,

$$u \pm v = \square_1 \\ u \pm w = \square_2 $$

In Aleksandr's and Joris' data, there were at least three solutions.

$$(a,b,c,d,w) = (16, 27, 12, 29, 263)$$

$$(a,b,c,d,w) = (19, 52, 16, 53, 177)$$

$$(a,b,c,d,w) = (29, 52, 8, 59, 369)$$

yielding the multi-grades valid for $k=(2,4,8)$,

$$16^k + 27^k + 11^k + 855^{k/2} = 12^k + 29^k + 15^k + 751^{k/2}$$

$$19^k + 52^k + 47^k + 1135^{k/2} = 16^k + 53^k + 43^k + 1495^{k/2}$$

$$29^k + 52^k + 19^k + 3319^{k/2} = 8^k + 59^k + 47^k + 1471^{k/2}$$

So, to recap, does $x_1^8+x_2^8+x_3^8+x_4^4 = y_1^8+y_2^8+y_3^8+y_4^4$ have infinitely many primitive solutions using this or some other parameterization? Or, if too difficult, what other $(a,b,c,d,w)$ will solve the third equation?