I was trying to prove Fermat's Last theorem by myself for the n=7 case for the first case via elementary methods

While trying to prove FLT for $n=7$, I came to find that there could be a certain class of solutions (counterfactual) for the counterfactual case that FLT is false for $n=7$:

$(d^7+x^7+y^7)^7 = (d^7+x^7-y^7)^7 + (d^7-x^7+y^7)^7$

where $x, y$ and $d$ are coprime and all of them are coprime to $7$. Meaning, this equation is a class of solutions that counterfactually satisfy the first case of FLT for $n=7$.

Could someone please prove the above equation has no solutions by starting from my equation using a elementary approach?

Here are some congruences I derived from the equation: $$\begin{aligned} 1. && 2d^{49} &\equiv (d^7 + 2y^7)^7 \pmod{x - y} \\ 2. && 2y^{49} &\equiv (2d^7 + y^7)^7 \pmod{d + x} \\ 3. && 2x^{49} &\equiv (2d^7 + x^7)^7 \pmod{d + y} \\ 4. && d &= x + y + 49m \text{ (which implies } d \equiv x + y \pmod{7}\text{)} \\ \end{aligned}$$

P.S. I mean if true, this is interesting because it means that a number must be capable of being written as the sum of three septics, and that means that if it is then the solutions must also be solutions of three septics.

It would have been easy to solve this had I been sure that $d+x$, $x+y$ or $x-y$ is prime and equal to $1 (\pmod 7)$ but sadly it is not the case.

I am also a second-year undergrad math student and since you all know so much, I respect your knowledge to guide me in the right direction.