To solve,

$$A^4+B^4 = C^4+D^4$$

we use Euler's method. Let,

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

and define $p = (a^3 - b),\, q = a y,\, r = b (a^3 - b),\, s = y.\,$ The equation above transforms to the simple form,

$$(a^3 - b) (b^3 - a) = y^2$$

I. Rational points

This is birationally equivalent to an elliptic curve. Assume the case $a=n.$ Six Seven "smallish" solutions are known (with $b_6$ and $b_7$ found by Seiji Tomita in this related post):

$b_1 =\frac{n\,(\color{red}{1})}{(1)}$

$b_2 =\frac{n\,(\color{red}{4} + n^2 + 10n^4 + n^6)}{(1 + 10n^2 + n^4 + 4n^6)}$

$b_3 =\frac{n\,(\color{red}{9} - 44n^2 + 190n^4 + 100n^6 + n^8)}{(1 + 100n^2 + 190n^4 - 44n^6 + 9n^8)}$

$b_4 =\frac{n\,(\color{red}{16} - 543n^2 + 4632n^4 + 15100n^6 + 10632n^8 + 22758n^{10} + 6568n^{12} + 5820n^{14} + 552n^{16} + n^{18})}{(1 + 552n^2 + 5820n^4 + 6568n^6 + 22758n^8 + 10632n^{10} + 15100n^{12} + 4632n^{14} - 543n^{16} + 16n^{18})}$

$b_5 =\frac{n\,(\color{red}{25} -3524n^2 + 113482n^4 + 979388n^6 +\,\dots\, + 45836n^{18} + 69418n^{20} +2092n^{22} + n^{24})}{(1 + 2092n^2 + 69418n^4 + 45836n^6 +\,\dots\, + 979388n^{18} + 113482n^{20} - 3524n^{22} + 25n^{24})}$

$b_6 = \frac{n(n^{38}+6234n^{36}+569433n^{34}-1574764n^{32}+\,\dots\,+32622105n^6+1538106n^4-15551n^2+\color{red}{36})}{(36n^{38}-15551n^{36}+1538106n^{34}+32622105n^{32}+\,\dots\,-1574764n^6+569433n^4+6234n^2+1)}$

$b_7 = \frac{n(n^{48}+15704n^{46}+3430692n^{44}-57632376n^{42}+\,\dots\,+603165288n^6+13866564n^4-54088n^2+\color{red}{49})}{(49n^{48}-54088n^{46}+13866564n^{44}+603165288n^{42}+\,\dots\,-57632376n^6+3430692n^4+15704n^2+1)}$

$b_8 = \;?$

A curious feature is the coefficients of the numerator and denominator are palindromic wrt to each other.

II. Identities

These points yield nice identities (after a change of variables) of symmetric form,

$$f(\alpha, \beta)^4 + f(\beta, -\alpha)^4 = f(\alpha, -\beta)^4 + f(\beta, \alpha)^4$$

with the smallest non-trivial $f(\alpha, \beta)$ being of degree $7$.

III. Updates

Update 1. As pointed out by Sidharth Ghoshal (when only six $b_m$ were known):

1. The coefficients of $b_m$ sum to $2^k$, namely $2^0,\,2^4,\,2^8,\,2^{16},\,2^{24},\,2^{36}.$ (Why?)
2. The degree $d$ of the denominators are $0, 6, 8, 18, 24, 38$.
3. He pointed out that it seems both the power $k$ and degree $d$ are functions of $m$.

$$\begin{array}{|c|c|c|c|} \hline m&m^2&k&d\\ \hline\color{blue} 1&1&0&0\\ \hline 2&4&4&6\\ \hline\color{blue} 3&9&8&8\\ \hline 4&16&16&18\\ \hline\color{blue} 5&25&24&24\\ \hline 6&36&36&38\\ \hline\color{blue} 7&49&48&48\\ \hline 8&64&64&66\\ \hline\color{blue} 9&81&80&80\\ \hline \end{array}$$

At the suggestion of Deyi Chen, the expression for $b_1$ has been made consistent with other $b_m$ for odd $m.\,$ Hopefully someone can find $b_8$ so we can complete this table. (Completed.)

Update 2. Thanks to prompt help from Seiji Tomita, we managed to find $b_8$. So what I labelled as $b_9$ earlier was indeed the case. Both fit the patterns above.

$b_8 = \frac{n(\color{red}{64} - 158335n^{2} + 91670880n^{4} + 7908319600n^{6} + \,\dots\,- 802597088n^{60} + 16598640n^{62} + 34976n^{64} + n^{66})}{(1 + 34976n^{2} + 16598640n^{4} - 802597088n^{6} + \,\dots\, + 7908319600n^{60} + 91670880n^{62} - 158335n^{64} + 64n^{66})}$

$b_9 = \frac{n(\color{red}{81} - 407672n^{2} + 482840284n^{4} + 77282464024n^{6} + \,\dots\, - 7601467144n^{74} + 67097980n^{76} + 70888n^{78} + n^{80})}{(1 + 70888n^{2} + 67097980n^{4} - 7601467144n^{6} + \,\dots\, + 77282464024n^{74} + 482840284n^{76} - 407672n^{78} + 81n^{80})}$

IV. Questions

1. Tomita and I found these $b_m$ using different techniques. For any given positive integer $m$, is it always possible to find a rational polynomial $b_m$ that fit the patterns in the table above, such as the palindromicity and numerator having a linear term that is $m^2$?
2. And how do we explain Ghoshal's observation that the coefficients sum to $2^{ 4\left\lfloor \frac{m^2}{4} \right\rfloor }$?

P.S. A related question was asked in this post but focuses on other aspects.