More elliptic curves for $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$?

Given a rational number $m$, let $C_{m}$ be the intersection of two quadrics given by (3) and (4) in the original question. The way to search for points is to test the curve $C_{m}$ to see if it has local points. I did this for all $m$ with height $\leq 193$ (since I knew I'd find points on $C_{193/18}$). There are only $18$ rational numbers for which $C_{m}$ has points in $\mathbb{R}$, and $\mathbb{Q}_{p}$ for $p = 2$ an all primes of bad reduction of the Jacobian of $C_{m}$. (I did this computation in Magma, and it was complicated by the fact that Magma currently has a bug in its IsLocallySolvable command.)

(The $18$ rationals of height less than or equal to $193$ for which $C_{m}$ has local points are $157/150$, $163/150$, $151/126$, $181/150$, $121/96$, $103/78$, $97/72$, $79/54$, $37/25$, $67/42$, $49/24$, $73/25$, $109/25$, $31/6$, $133/25$, $169/25$, $181/25$, and $193/18$.)

Searching for points on these $C_{m}$ of height up to $10^{8}$ reveals points for $m = 157/150$, $m = 37/25$, $m = 31/6$ and the point on $m = 193/18$ that we already knew about. New solutions to (1) that arise from these points are \begin{align*} a &= 841263, b = -792940, c = 3852350, d = -44410\\ a &= 35847220, b = -34122866, c = 53902630, d = 2542025\\ a &= 39913670, b = -23859495, c = 15187700, d = 10116014,\\ \end{align*} which come from $m = 157/150$, $m = 37/25$ and $m = 31/6$, respectively. (I actually found two points on the $m = 31/6$ curve. The other point gives a solution to (1) which is a permutation of the $a = 35847220$ solution.)

It may be possible to prove that $C_{m}$ has no points on it for other choices of $m$ using the fact that $C_{m}$ is a four-cover of its Jacobian and using other techniques (the Cassels-Tate pairing, 3-descents, $L$-functions) to show that $C_{m}$ represents an element of Sha.

Seiji Tomita found a fifth elliptic curve $E$ as $m = \frac{19^2}{61}$ with an explicit solution. In summary, there are now seven "small" primitive solutions to,

$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4 = z^4$$

or equivalently,

$$(p-2q + r)^4 + (p-2q - r)^4 + (q + s)^4 + (q - s)^4 = (2p - 2q)^4$$

with $z$ less than 60 million:

$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline E&m&a&b&c&d&z&\text{by}&\text{year}\\ 1&\color{blue}{\frac{31^2}{61}}&5400& 1770& -2634& 955& 5491& \text{Brudno}&1964\\ 2&\frac{211}{175}&-31764& 7590& 27385& 48150& 51361& \text{Wroblewski}&2005\\ 3&\color{blue}{\frac{19^2}{61}}&1984340& -107110& 1229559& -1022230& 2084559& \text{Tomita}&2015\\ 3&\color{blue}{\frac{19^2}{61}}&3597130& -1953890& 561760& 1493309& 3698309& \text{Tomita}&2015\\ 4&\frac{157}{150}&841263& -792940& -44410& 3852350& 3856263& \text{Rouse}&2014\\ 5&\frac{31}{6}&39913670& -23859495& 15187700& 10116014& 41357889& \text{Rouse}&2014 \\ 5&\frac{31}{6}&53902630& 2542025& 35847220& -34122866& 58169009& \text{Rouse}&2014\\ \hline \end{array}$$

The variable $m$ can be recovered as,

$$m =\frac{3q^2+s^2}{p^2-r^2}$$

(I have no idea if $\displaystyle\frac{19^2}{61}$ and $\displaystyle\frac{31^2}{61}$ is just coincidence.) Incidentally, as pointed out by Rouse, if $m$ is solvable, then so is $m'=\frac{m+1}{m-1}$. Hence, Tomita's $m = \frac{211}{150}$ in his website yields $m' = \frac{19^2}{61}$ cited here.