As a background, it has been established that solutions to,
$$a^4+b^4+c^4+d^4 = e^4\tag1$$
must have either $d^2-e^2$ or $d^2+e^2$ divisible by $5^4$. Furthermore, if $a=0$, it was proven by Ward that either $-d+e$ or $d+e$ is divisible by $2^{10}$. The smallest is, $$95800^4+ 414560^4+ 217519^4 = 422481^4$$ and in fact we have $d+e = 217519 + 422481 =2^{10}\times5^4$. For
$$a^4+2b^4 = c^4+4d^4\tag2$$
the only unscaled and non-zero solution is by Elsenhans and Jahnel as, $$1484801^4+2\times 1203120^4= 1169407^4+4\times 1157520^4$$ though presumably there should be infinitely many. On a hunch, I checked it for similar congruences and sure enough, $$a+c = 1484801 + 1169407 = 2^{15}\times3^4$$
Q: Is it true that solutions to $(2)$, similar to Ward's constraint, must have $a\pm c$ divisible by $2^{10}$ or even $2^{15}$?
P.S. I had emailed Elsenhans and Jahnel about this question but they are unsure as to the answer.