In this case, it pays to work on the curve $E^\prime$ that is 2-isogenous to $E$, which is given by the equation

$$y^2 = x^3 + 404100192598226941365253x^2+ 1470175712258164849983363482095324897635296971x.$$

It is relatively easy to find the 7 independent points

$(110776963853866550724016 : -80505869468630089210131377497504980 : 1)$,

$(32748343004768539696321 : 22729843039610504103217516288626615 : 1)$,

$(33849219894746702769856 : -23485175376256902691040244680173080 : 1)$,

$(83632124830832490757371607899/221801449 : 2583856339678270035941408483841167567865756495/3303288979957 : 1)$,

$(708467443931328192488072059/9 : 18905656692867793914636996097237707997855/27 : 1)$,

$(82669556642133426513025 : -58721781791249872164660769545564855 : 1)$,

$(15653787556726119039025946369377024/458311398169 : 7352032091260403572244890413120572309377214021131920/310270858512236803 : 1)$

on $E^\prime$.

To find the final generator for $E^\prime$, we need to apply 4-descent. We can find the final point by applying 4-descent to any number of 2-covers of E'. I got a hit on the second one I tried, which was given by

$$C: y^2 = 40629834885797531781124x^4 + 168889375516514838017136x^3 - 522073637939703266788148x^2 - 1468041555233000587306308x + 3129498870121696083297961.$$

One 4-coverings of C is given by the intersection of the quadrics

$$26478x^2 + 149391xy + 147873xz + 592534xw - 114021y^2 - 58434yz + 336829yw + 118629z^2 + 510074zw + 438488w^2$$

and

$$148610x^2 + 19042xy + 1022361xz - 1631065xw - 112833y^2 + 39500yz + 44513yw - 182441z^2 + 822710zw + 972880w^2$$

in $\mathbb{P}^3$. This has a rational point at $(9608:-18440:7168:6485)$.

The corresponding rational point $Q$ on $E^\prime$ is given by

$(-1117913472469704108682566452343804675555656978809017451310789/5063218268474760928272238797769806961 : 1057252852298703272836151916049253048501046484070626117498677044566639368499019825133763515/11393049242085218620499121263510902732524161971013459959 : 1)$,

which has canonical height $106.35$. Using the other generators, we can find the somewhat smaller point $Q^\prime$

$(-101978171213582065261997068234175010796855613305889831/9689020674223383052192046420224 : 159945556170312399327355677945959801245147861170127803803586963373362965000839965/30159200458008252634403235853082432918163525632 : 1)$

having canonical height $91.18$ that is independent of the first seven generators.

The discriminant of $E^\prime$ is somewhat smaller than that of $E$, so arguably $E^\prime$ is the model that you'd want to present. However, if you want to work on $E$, then we can map the generators of $E^\prime$ to $E$. A "nice" choice of points that generator $E(\mathbb{Q})$ modulo torsion is:

$(225814423074482435222996 : -34626227198383025293628840137707870 : 1)$,

$(-151681843950496144133344 : 94669696550541536347147233612637650 : 1)$,

$(-209328450874403020690564 : 82198277984719522705484038947202890 : 1)$,

$(30089244387022819458738989/4 : -164963185085547559044466862956833045285/8 : 1)$,

$(1521645572343712794396956 : -1853309164504557572407788096216777750 : 1)$,

$(-108427296385410370841188 : 92566828570027620341222482486229910 : 1)$,

$(50259473549510044315364816963047/529 : -356309090633909063646983926957844863796356762164/12167 : 1)$,

$(-825240948184709245504424305468714422091334601593572177730440863632644222609685944791784/3476235906470724630424489414741262801265768781870915852371028441: 13350668242169750290895245364589704907786881444602849078579635591912831036781435950810619013243973916848136198892088368285547068910/204957522013550460549477456089656119391219129674638303351158680324230445521351939753256298608739 : 1)$.

The final generator has canonical height $169.76$, which means that it would be more difficult to find by searching directly on 4-coverings of $E$.

In this case, it pays to work on the curve $E'$ that is 2-isogenous to $E$, which is given by the equation $$ y^2 = x^3 + 404100192598226941365253x^2+ 1470175712258164849983363482095324897635296971x. $$

It is relatively easy to find the 7 independent points

(110776963853866550724016 : -80505869468630089210131377497504980 : 1),
(32748343004768539696321 : 22729843039610504103217516288626615 : 1),
(33849219894746702769856 : -23485175376256902691040244680173080 : 1),
(83632124830832490757371607899/221801449 : 2583856339678270035941408483841167567865756495/3303288979957 : 1),
(708467443931328192488072059/9 : 18905656692867793914636996097237707997855/27 : 1),
(82669556642133426513025 : -58721781791249872164660769545564855 : 1),
(15653787556726119039025946369377024/458311398169 : 7352032091260403572244890413120572309377214021131920/310270858512236803 : 1)

on $E'$.

To find the final generator for $E'$, we need to apply 4-descent. We can find the final point by applying 4-descent to any number of 2-covers of $E'$. I got a hit on the second one I tried, which was given by $$ C: y^2 = 40629834885797531781124x^4 + 168889375516514838017136x^3 - 522073637939703266788148x^2 - 1468041555233000587306308x + 3129498870121696083297961. $$

One 4-covering of $C$ is given by the intersection of the quadrics $$ 26478x^2 + 149391xy + 147873xz + 592534xw - 114021y^2 - 58434yz + 336829yw + 118629z^2 + 510074zw + 438488w^2 $$ and $$ 148610x^2 + 19042xy + 1022361xz - 1631065xw - 112833y^2 + 39500yz + 44513yw - 182441z^2 + 822710zw + 972880w^2 $$ in $\mathbb{P}^3$. This has a rational point at (9608:-18440:7168:6485).

The corresponding rational point $Q$ on $E'$ is given by

(-1117913472469704108682566452343804675555656978809017451310789/5063218268474760928272238797769806961 : 1057252852298703272836151916049253048501046484070626117498677044566639368499019825133763515/11393049242085218620499121263510902732524161971013459959 : 1),

which has canonical height $106.35$. Using the other generators, we can find the somewhat smaller point $Q'$

(-101978171213582065261997068234175010796855613305889831/9689020674223383052192046420224 : 159945556170312399327355677945959801245147861170127803803586963373362965000839965/30159200458008252634403235853082432918163525632 : 1)

having canonical height $91.18$ that is independent of the first seven generators.

The discriminant of $E'$ is somewhat smaller than that of $E$, so arguably $E'$ is the model that you'd want to present. However, if you want to work on $E$, then we can map the generators of $E'$ to $E$. A "nice" choice of points that generator $E(\mathbb{Q})$ modulo torsion is:

(225814423074482435222996 : -34626227198383025293628840137707870 : 1),
(-151681843950496144133344 : 94669696550541536347147233612637650 : 1),
(-209328450874403020690564 : 82198277984719522705484038947202890 : 1),
(30089244387022819458738989/4 : -164963185085547559044466862956833045285/8 : 1),
(1521645572343712794396956 : -1853309164504557572407788096216777750 : 1),
(-108427296385410370841188 : 92566828570027620341222482486229910 : 1),
(50259473549510044315364816963047/529 : -356309090633909063646983926957844863796356762164/12167 : 1),
(-825240948184709245504424305468714422091334601593572177730440863632644222609685944791784/3476235906470724630424489414741262801265768781870915852371028441 : 13350668242169750290895245364589704907786881444602849078579635591912831036781435950810619013243973916848136198892088368285547068910/204957522013550460549477456089656119391219129674638303351158680324230445521351939753256298608739 : 1).

The final generator has canonical height $169.76$, which means that it would be more difficult to find by searching directly on 4-coverings of $E$.