Let us fix some notations, that slightly differ from the OP-ones.
Please compare with Tito's detailed question MSE 4857229. (I keep notations from my answer in there.)
We use capital letters $X,Y,Z$ for a rational solution of
$$
\tag1
X^4 + Y^4 + Z^4 = 1\ .
$$
Below, i will usually sort the values so that $0<X\le Y\le Z$.
Let $t$ be the common denominator of $X,Y,Z$, then we obtain a solution for
$$
\tag2
x^4 + y^4 + z^4 = t^4\ .
$$
Let $g$ be the slightly asymmetric function
$$
\tag3
g(X,Y,Z) =
\frac{(X-Y)^2 - Z^2 - 1}{X^2 - XY + Y^2 + (X-Y)}\ ,
$$
which appears in the OP-equations $(2)$, $(3)$, $(4)$.
Let $f$ be the function
$$
\tag4
f(u,v) =
\begin{bmatrix}
u^4 & u^3 & u^2 & u & 1
\end{bmatrix}
\begin{bmatrix}
1 & -4 & 0 & -8 & 4 \\
-4 & 8 & -16 & 48 & -16 \\
0 & -16 & 48 & -64 & 0 \\
-8 & 48 & -64 & 32 & 32 \\
4 & -16 & 0 & 32 & -48
\end{bmatrix}
\begin{bmatrix}
v^4 \\ v^3 \\ v^2 \\ v \\ 1
\end{bmatrix}
\ .
$$
Then MSE 4857229 offers us the following structural connections:
Start with a solution $(X,Y,Z)$ of the Euler equation $X^4+Y^4+Z^4=1$. Associate:
$$
\begin{aligned}
u &= g(X,Y,Z)\ ,\\
v &= g(Y,Z,X)\ ,\\
w &= g(Z,X,Y)\ ,
\end{aligned}
$$
as in the OP-equations $(2)$, $(3)$, $(4)$. Then $u,v,w$ satisfy
$$
\tag5
2(u+v+w)-uvw-4=0\ .
$$
The important point is now the following:
Lemma A: $v$ lifts to a point $(d,v)$ on the hyperelliptic curve with affine equation in the $(D,V)$-world:
$$
C_u\ :\qquad D^2 = f(u,V)\ .\\
$$
See MSE 4857229 for details on the lift.
The R.H.S. above is a polynomial of degree four in $V$, so we are in the genus one case.
OP explicitly did a good work in collecting all these details,
and we have now from the point of view of a small (experimental) research project the question on the search of "good" values $u$ (of small height), so that (definition of "good") there is a rational point on $C_u$.
The following ideas are natural, and were already used in the many related posts.
$(A)$ The $S_3$-action mule: If $(u,v,w)$ is a good tuple, then each tuple $(u', v', w')$ obtained by permuting the components is also a good tuple.
$(A')$ The $\pm^3$-mule. We start with $(u,v,w)$, obtain a solution $(X,Y,Z)$ for $(1)$, change the sign in the components and pick one new solution
$(X',Y',Z')=(\pm X,\pm Y,\pm Z)$, and associate the triple $(u',v',w')$ for this one. This looks trivial, but is not in applications. For instance, for a "good" $u$ as in the triple we started with, we know corresponding $v$-points $g(Y,Z,X)$ (which is the second component of the triple), but then also $g(-X,Z,-Y)$, $g(Y,-Z,X)$, $g(-X,-Z,-Y)$ without effort because of the $S_3$-mule $(A)$, and because $u$ can be realized as
$$
u = g(X,Y,\pm Z)=g(-Y,-X,\pm Z)\ .
$$
$(B)$ The "cheetah" method: We look in the list of known solutions, for instance oeis.org/A003828, or Tito's long list in MSE 1853223, and for each solution $(x,y,z,t)$ (in oeis displayed as $(a,b,c,k)$), we simply associate $u,v,w$ and look for values of small(er) height among them.
$(C)$ EC-jumps: We start with one known pair $(u,v)$, associate the curve $C_u$ of equation $D^2=f(u,V)$, then the elliptic curve $E_u$ for $C_u$ with a birational map $\beta:E_u\to C_u$, essentially using the known rational value $v$. The curve $E_u$ may have positive rank, knowing it and/or at least some points in $E_u(\Bbb Q)$ of infinite order, we may send them back via $\beta$ to $C_u$, and for each value $v'$ obtained so we associate a new triple $(u,v',w')$ with $w'$ as in $(5)$. This step depends on the complexity of $E_u$, and on the success of our descent strategy.
$(D)$ Brute-force: This is excluded in the question, but may still be useful in combination with the other methods. A good version of it is to use the pari/gp functionality of hyperellratpoints (due to Michael Stoll, Bill Allombert) to get points of small naive height on $C_u$.
$(E)$ WC-triviality: We start with a point $u$, known to be good or not. We build $C_u$, and search more structurally for a rational point on it. Associate its Jacobian $E_u$, and try to get $C_u$ identified with some Selmer group element $\xi_u$ for $E_u$. Then information on and computations in some Selmer group for $E_u$ ideally leads to a rational point, if $\xi_u$ is trivial.
Examples:
$(A)$ and $(B)$:
We start with the solution $(x,y,z,t)=(95800, 217519, 414560, 422481)$, Roger Frye, 1988, and build $(X,Y,Z)=\frac 1t(x,y,z)$. Then the associated $(u,v,w)$ triples are:
$$
\begin{aligned}
(X,Y,Z) &\to (u,v,w)=\left(\
\frac{1000}{47}\ ,\
-\frac{1041}{320}\ ,\
-\frac9{20}\
\right)\ ,
\\
(Y,X,Z) &\to (u',v',w')=\left(\
-\frac{167767337}{43538900}\ ,\
-\frac{6899820729}{369596780}\ ,\
-\frac{71490240}{101943281}\ ,
\right)\ ,
\end{aligned}
$$
and each of the values in the list $u,v,w;u',v',w'$ is a good value.
Observe that the small assymetry in $g$ leads to small values for $(u,v,w)$, but to bigger ones for $(u',v',w')$. So the $S_3$-mule depends on the sign of the wind.
$(A)$ at the level of the associated elliptic curve.
To see a further aspect of the $S_3$-mule, it is maybe good to take a look at numeric data. We have seen above which triple $(u,v, w)$ corresponds to Frey's solution $(X,Y,Z)$. The "easy" value $-9/20$ is on position $w$, the next easy value on position $u$. But for each permutation we obtain an elliptic curve, and here is the list:
- Starting from $(X,Y,Z)$, $E_u$ is:
$$
\small
y^2 = x^3 - x^2 + 42203431924184331476428x
+ 19708860022001754313999285183771968
\ .
$$With some effort we find on this curve the (integral) point:
$$
\small
(189408782348, \ 185735647714986000)\ .
$$
- Starting from $(X,Z,Y)$, the associated elliptic curve is $E_u$:
$$
\small
y^2 = x^3 + 7800804106591240085872347270947659749926440504618278762158577297827090363903521 x + 37530455728534262347322433701721255609064620464175305988049123583446684338461583586305368948564333841480231349206181922\ ,
$$ and with the same effort we find the (integral) point
$$
\small
(2941922994430820231717552531581784132321,\
293158493159260244914142554914180294608413653969352427026482)
$$ on it.
- Starting from $(Y,Z,X)$, the associated elliptic curve is $E_u$:
$$
\small y^2 = x^3 + 13945360743557326851876481x + 33681945402013610389723194465125500802
$$ and with the same effort we find now two points
$$
\small
(3767422235201,\ 11819171000608361922)\ ,\\
\small
(7046841826529451479/316969\ , 18996782644715674941660835536/178453547)
$$ on it. (The first one is an integral point.)
- Starting from $(Y,X,Z)$, the associated elliptic curve is $E_u$:
$$
\small
y^2 = x^3 - x^2 + 4479506503717291186854212434802788023482795873427058374070419477388x + 7572748078333402630994645782051181039196205097959115377969415435638899110817843915712509426488194692
$$ and with the same effort we find the (integral) point
$$
\small
(2190303762448735558129556967770828,\ 167009156212564330197459465965640919033939683393618)
$$ on it.
- Starting from $(Z,X,Y)$, the associated elliptic curve is $E_u$:
$$\small
y^2 = x^3 + 2265722465761 x - 3154189403034549278
$$ and with less effort we find the three points
$$\small
(1237921,\ 1244044242)\ ,\\
\small
(47971729/49,\ 16603172706/343),\ \\
\small
(37313463163849/15896169,\ 246260871807590439226/63378025803)
$$ on it. (First point is integral.) Because of the simple nature
of the case, we will use it again below.
- Starting from $(Z,Y,X)$, the associated elliptic curve is $E_u$:
$$\small
y^2 = x^3 + 160509683055190727130612845899398203252291734337571536313314064001 x - 34115152005914438925415391663420789399000309316750390099863501440778785102995727125525739887231362
$$ and with the same effort we find the point
$$\small
(318296637885530436525646003650241,\
7015842647868172123278010543581378874606005636480)
$$ on it.
So there are differences from case to case from the practical point of view.
$(A')$ for the Frey's solution
$(x,y,z,t)=(95800, 217519, 414560, 422481)$ again.
We change the sign of $X$ and obtain the new triple
$\left(\frac{29957400}{6538471}\ ,\ -\frac{1041}{320}\ ,\ \frac{52463}{660460}\right)
$.
Changing $Y\to -Y$ only we obtain $\left(-\frac{4209}{3500}\ ,\ \frac{30080}{6007}\ ,\ -\frac{9}{20}\right)$, we will see below this again.
Changing $Z\to -Z$ only we obtain $
\left(\frac{1000}{47}\ ,\ \frac{3521543}{9580960}\ ,\ \frac{330353}{48940}\right)$.
Changing two signs, for $X,Y$, we obtain $\left(-\frac{167767337}{43538900}\ ,\ \frac{30080}{6007}\ ,\ \frac{52463}{660460}\right)$.
Changing two signs, for $Y,Z$, we obtain $\left(-\frac{4209}{3500}\ ,\ -\frac{71490240}{101943281}\ ,\ \frac{330353}{48940}\right)$.
Changing two signs, for $Z,X$, we obtain $\left(\frac{29957400}{6538471}\ ,\ \frac{3521543}{9580960}\ ,\ -\frac{6899820729}{369596780}\right)$.
Changing all three signs, the new triple is the most complicated one, $\left(-\frac{167767337}{43538900}\ ,\ -\frac{71490240}{101943281}\ ,\ -\frac{6899820729}{369596780}\right)$.
$(C)$
We start with the simplest $u,v$ values from above, explicitly $u=-9/20$, $v=1000/47$. There $w=-\frac{1041}{320}$, and the two solutions of the system with OP-equations $(2)$, $(3)$, and $(4)$ are
$$
\begin{aligned}
& 95800, \ 217519,\ 414560,\ 422481\ ,\\
& 50237800, \ 632671960,\ 1670617271,\ 1679142729\ .
\end{aligned}
$$
The curve $C_u$ with equation $D^2=f(u,V)$ is explicitly:
$$
C_u\ :\ D^2 =
\frac{1280881}{160000} V^{4} - \frac{1669321}{40000} V^{3} + \frac{19989}{500} V^{2} - \frac{1241}{20000} V - \frac{2431119}{40000}\ .
$$
We know the point $(D_0,V_0)=(q,v)=\left(\frac{495260031}{441800}, \ \frac{1000}{47}\right)$.
The related elliptic curve is as mentioned:
$$
E_u\ :\ y^2 = x^3 + 2265722465761x - 3154189403034549278\ ,
$$
or also $y^2=(x - 978559)(x^2 + 978559x + 3223300182242)$,
it has thus a $2$-torsion point $T=(978559,0)$, and furthermore rank three, a choice of the generators being as mentioned:
$$
\begin{aligned}
P_1 &= (1237921, 1244044242)\ ,\\
P_2 &= (47971729/49, 16603172706/343)\ ,\\
P_3 &= (37313463163849/15896169, 246260871807590439226/63378025803)\ .
\end{aligned}
$$
The torsion-points $O,T$ are not directly useful for our purposes,
so let us take $P_1\in E_u(\Bbb Q)$, and move it back to a point on $C_u$. This point is explicitly
$$
(d, v') := \beta(P_1)=
\left(
-\frac{6507898503}{700000000}\ ,\
-\frac{4209}{3500}
\right)
\ .
$$
So we have a new pair, $(u,v_{P_1})=(u, v')=(u',v')=\left(-\frac 9{20}\ ,\ -\frac{4209}{3500}\right)
$,
then associate the corresponding $w'=w'_{P_1}=
\frac{30080}{6007}$, and the system with OP-equations $(2)$, $(3)$, and $(4)$ for the new triple,
$$
\left\{
\begin{aligned}
g(X',Y',Z') &=u'=u\ ,\\
g(Y',Z',X') &=v'\ ,\\
g(Z',X',Y') &=w'\ ,
\end{aligned}
\right.
$$ has two solutions $(X',Y',Z')$, which are leading to $(x',y',z',t')$-solutions:
$$
\begin{aligned}
& 95800, \ 217519,\ 414560,\ 422481\ ,\\
& 7592431981391,\ 22495595284040,\ 27239791692640,\ 29999857938609\ .
\end{aligned}
$$
The first one was known.
What about using $P_1+T$ instead of $P_1$?
Well, we then obtain the value $v''=v_{P_1+T}=w'=\frac{30080}{6007}$, and the associated $w$-value is $w''=w_{P_1+T}=v'=-\frac{4209}{3500}$.
The corresponding solutions $(x'',y'',z'',t'')$ are the same (after rearranging).
So there is some symmetry that should be isolated structurally.
What about using $P_2$ (instead of $P_1$)?
We obtain $v_{P_2}=-\frac{29393447736}{9584944225}$, the corresponding $w$-value is
$w_{p_2}=14486729065/814087256$, and the solutions for the triple $(u, v_{P_2},w_{P_2})$ are
$$\small
\begin{aligned}
&
59421842165791512201169,\
440517744543240750721000,\
1165970778032514255823760,\
1171867103503245199920081,\\
&21710111037730547416,\
54488888702794271560,\
99569174129827461335,\
101783028910511968041,
\end{aligned}
$$
so we land far away from the starting point. We have started with solution #1 in Tito's table from loc. cit., and obtain his solution #51, and Tomita's #71 in the same table!
Using the point of higher height $P_3$ we land here:
$$\tiny
\begin{aligned}
&
4906600101468927788661521948711588613529046335,\
32333025303302398783385143300102444875640407784,\
32920720418680293851782448531772814736860986760,\
38807308867109469306956130726405668929338587841,\\
&2496156394991961814239989060375233511277443800,\
5869555596785291641235700962130096065667529360,\
7620654701590271369073375691220522430753106729,\
8234762698423310578177015206662106449800564521 \ .
\end{aligned}
$$
$(D)$ Brute force is excluded in the question, but sometimes it lets us spare computing effort. We can ask pari/gp for rational points on the hyperelliptic curve $D^2=f(u,V)$, seen in the parameters $(D,V)$. For instance if we know that $u=-9/20$ is "good", which $v$-values are matching it? Sage-code for this:
sage: R.<V> = PolynomialRing(QQ)
sage: F = 1280881/160000*V^4 - 1669321/40000*V^3 + 19989/500*V^2 - 1241/20000*V - 2431119/40000
sage: pari(F).hyperellratpoints(20000, 1)
[[1000/47, 495260031/441800]]
And hyperellratpoints immediately finds the above point, then stops because of the flag one. The corresponding pari code is also simple:
? F = 1280881/160000*V^4 - 1669321/40000*V^3 + 19989/500*V^2 - 1241/20000*V - 2431119/40000;
? hyperellratpoints(F, 20000, 1)
%7 = [[1000/47, 495260031/441800]]
$(D)$ again. There is a "good" $u$-value, Bremmer's $u=-5/44$, and we would like to find an associated $v$. One good value that works is in the triple
$$
(u,v,w)=
\left(
-\frac 5{44}\ ,\
\frac{57878913}{12642040}\ ,\
-\frac{2741924904}{1401894085}
\right)\ ,
$$
and starting from $u$ with an effort like 2\cdot 10^6$ (for the computer, not for us) is not finding any solutions. So brute force is not an option.
$(D)+(C)+(A')$ for $u=-\frac5{44}$. Is there any smaller $v$-value that works for this $u$? (Smaller means here of smaller height then the known value of Bremmer $57878913/12642040$.)
We take the above pair, and associate the corresponding elliptic curve $E_u$,
$$
y^2 = x^3 - x^2 + 349942184229228 x - 11167797929528591502588\ ,
$$
or also $y^2=(x^2 + 17248665x + 647458645760118)(x - 17248666)$.
Its equation still has coefficients that fit in the line, and we know enough rational points on it because of $(A')$! Any "better" $v$ on $C_u$ leads to a rational point on $E_u$ of moderate height. It turns out that $E_u$ has rank three, it has the torsion point $T=(17248666, 0)$, and a basis is
$$
\small
\begin{aligned}
Q_1 &= (17487516, 17312851170)\ ,\\
Q_2 &= \left(\frac{133466957657960755098}{1953596039521}\ ,\ \frac{1572415798225425554647417475088}{2730562673994936431}\right)\ ,\\
Q_3 &= \left(\frac{119965710178645015353945839868}{9789745694666738761}\ ,\ \frac{41551451427975081812132871307757363563736682}{30630713669485435307521202491}\right)\ .
\end{aligned}
$$
We built all linear combinations of the points $Q_1,Q_2,Q_3,T$ with integer coefficients (up to five, say), and plot only those with smaller values of the height. Here are they:
57878913/12642040
-2741924904/1401894085
4669000304/944254963
-43000836761/25579904000
-117620301817/53219719132
-343268956016/144380152505
-580735975701/314467586696
8541935778968/2036027368195
-10366183618512/3739167824077
28531188247669/5494554320180
So the $v$-value we started with is "good enough".
This is also supported by the height/complexity of the points on the associated curve $E_u$.
$(E)$ Here is a case where we start with a $u$-value, Bremmer's
$$
u= -\frac 3{40}\ ,
$$
and have no idea if a rational $v$ can be found on the curve $C_u$ with equation $D^2 =f(u,V)$. What can be done? First of all, we try a brute force search. No results.
We associate $C_u$, then for it the Jacobian, which is an elliptic curve. We need its equation...
sage: R.<D,V> = PolynomialRing(QQ)
sage: Cu = Curve(-D^2 + f(-3/40, V))
sage: J = Jacobian(Cu)
sage: J
Elliptic Curve defined by y^2 = x^3 + 140719990850401/409600000000*x - 1506338457231274914479/131072000000000000 over Rational Field
sage: J.minimal_model()
Elliptic Curve defined by y^2 = x^3 + 140719990850401*x - 3012676914462549828958 over Rational Field
And now we should try to understand the above elliptic curve and how $C_u$ is a torsor for it... However, if there is a rational point $v$, then we expect (in the sense of $(A')$) many rational points on the jacobian $J$. So a good first check, also a good idea when searching for good $u$-points, is to check if this Jacobian of $C_u$ has some "expected" rank three or more... I started of course a search for the rank, but have to wait. At any rate, to have a practical answer the question stated about how to find further small "good" $u$-values, this may be one. Compute Jac$(C_u)$ as an elliptic curve and try to get its rank, consider those values that lead to a rank three or more.
We can find for instance on the above minimal model some points...
sage: E = J.minimal_model()
sage: E.simon_two_descent(lim1=600, lim3=2000)
(and there is some deprecation warning telling me to ask for the rank with the pari-algorithm... but finally)
(2, 5, [(11327441 : 5895324498 : 1)])
So the rank is between $2$ and $5$, and we have found only one point.
(Even if we get rank three, it is still unclear that our torsor is trivial, we have to compute...) The computation is checking in between some quartics. Here is the final part of the monolog of the interpreter started with...
E.simon_two_descent(lim1=Integer(600), lim3=Integer(2000), verbose=Integer(2))
It concludes (only)
#III(E/Q)[2] <= 8
#E(Q)[2] = 2
#E(Q)/2E(Q) >= 4
1 <= rank <= 4
!!! III should be a square !!!
hence
[E(Q):phi'(E'(Q))] >= 8
#S^(phi')(E'/Q) = 32
#III(E'/Q)[phi'] <= 4
#S^(2)(E/Q) = 32
#III(E/Q)[2] <= 4
#E(Q)[2] = 2
#E(Q)/2E(Q) >= 8
2 <= rank <= 4
reduced generators = [[66402, 5895324498]]
points = [[0, 0], [66402, 5895324498]]
v = [2, 5, [[11261039, 0], [11327441, 5895324498]]]
(2, 5, [(11327441 : 5895324498 : 1)])
(... have to stop here ...)
