Numbers greater than $26033514998417$ that can be expressed as a sum of two positive 4th powers in more than 1 way

Question

Are there any numbers greater than $26033514998417$ which can be expressed as a sum of two positive 4th powers in atleast 2 ways ?

And if there are any such numbers, then, please include the number(s) that u have found in your answer.

In case, if anybody wants to know how the number

$26033514998417$ can be expressed as a sum of two positive 4th powers in 2 ways, here it is: https://math.stackexchange.com/a/4237040/857041

Answer 1

$$17472238301875630082 = 62047^4 + 40351^4 = 59693^4 + 46747^4$$ from A018786, all primes btw.

Dunno why that list stops at 835279626752, though, presumably because a sequence must be complete, i.e. all numbers smaller and not in the list must not have the property.

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Addendum: and here is a list with 516 primitive entries from Daniel J. Bernstein (found via A003824), that ends

...
955378 168531 883693 688026
960319  87924 942337 498984
962724 455837 870557 756684
972841 283961 967979 394717
988582 158157 976563 463702
989216 283949 958163 590024
989426 503473 938698 704399
989727 161299 913141 717447
990518   1039 967823 540326

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Addendum 2: Here is a 146-digit solution from PrimePuzzles 103

75326517042882955509049316560407015204559148957492168402274838923500008575948738239692316743036018051328508933923770491774189595934532956237348642
= 1679539956802461427023806692932554869^4
+ 2864939822128245005298014063613916133^4
=  735636962517662175684582548040182073^4
+ 2096549864621014042130013452441703601^4

They mention an identity from a collection by Edward Brisse on EulerNet: $$f_2^4(a,b)+f_2^4(b,-a)=f_2^4(a,-b)+f_2^4(b,a)$$ where $$f_2(a,b)= -a^{13}+a^{12}b+a^{11}b^2+5a^{10}b^3+6a^9b^4-12a^8b^5-4a^7b^6+7a^6b^7-3a^5b^8-3a^4b^9+4a^3b^{10}+2a^2b^{11}-ab^{12}+b^{13}$$

The collection has also (other) identidies of degree 7, 13, 19 and 31 for $a$, $b$, $c$, $d$. And there are identities that solve $x^4+y^4+z^4=2w^4$ like

$$(a^2-b^2)^4 + (a^2 + 2ab)^4 + (2ab + b^2)^4 = 2(a^2 + ab + b^2)^4$$

So this rabbit hole seems to go infinitly deep, both in terms of formulae and in terms of solutions you can generate with each formula.