As we all know that Sophie-Germain equation says that $x^4 + 4y^4$ is factorizable. Now, I tried to extend this definition further more to a general case as follows: $$x^{2^{4n-2}}+ 2^{4n-2} y^{2^{4n-2}} = (x^{2^{4n-3}}+2^{2n-1} y^{2^{4n-3}}+ 2^n {(xy)}^{2^{4n-4}})(x^{2^{4n-3}} + 2^{2n-1} y^{2^{4n-3}}- 2^n (xy)^{^{4n-4}}) \quad (1)$$ The above factorability can be easily proven. Now, firstly we consider $n = 1$ equation ($1$). $$x^4 + 4y^{4} = (x^{2} + 2y^{2} - 2(xy))(x^{2} + 2y^{2} + 2(xy)) \qquad (2)$$ Equation ($2$) is simply the Sophie-Germain identity's statement.
Now, equation ($1$) can be re-written as follows: $$(x^2)^2 + (2y^2)^2 = (x^2 + 2y^2 - 2(xy))(x^2 + 2y^2 + 2(xy)) \qquad (3)$$
It can be easily shown that on the Right-hand side of equation ($3$) one can never find perfect square for any $x,y$ belonging to $N$. The reason is $\gcd$ of both the factors on the RHS is always one which implies that it has no factors in common.
Now, a number is a perfect square iff it's prime pairs are equal, e.g $4 = 2*2, 36 = 2*2 * 3 * 3 = 2*3*2*3$. But in above case of equation ($3$) the $\gcd$ of product pairs is always one which implies that they have no prime pairs in common because they have no divisors in common. Hence, $$(x^2)^2 +(2y^2)^2 \neq (c^2)^2$$
Now, by using equation ($1$) and the arguments of $\gcd$ used above can be extended for any $n \in \mathbb{N}$ as follows:
$$x^{2^{4n-2}}+ 2^{4n-2} y^{2^{4n-2}} = (x^{2^{4n-3}})^{2}+ (2^{2n-1} y^{2^{4n-3}})^{2} \qquad (4)$$ Now, equation($4$) is also factorable into factors whose $\gcd$ is always $1$ as follows from ($1$).
Hence, can we say that, in equation ($4$) for any $n \in \mathbb{N}$, the Fermat's last theorem is true.
Your proof looks great! If you want to see more about this particular topic, here's a paper about Sophie Germain's work on proving Fermat's Last Theorem. I've also edited your proof to make it more readable, check it out.