[Math] Question about (probably false) elementary proof of Fermat’s Last Theorem

Can somebody explain to me what is wrong with the following argument for a proof of Fermat's Last Theorem?

Suppose somewhere the following theorem has been proven:

Theorem:

If:

$\quad \quad \quad \quad x^p = y^p + z^p$

$\quad \quad \quad \quad \gcd(x,y,z) = 1$

$\quad \quad \quad \quad p \gt 2$ and prime

Then:

$ \quad \quad \quad \quad x – y = r^p$,

$ \quad \quad \quad \quad (x^p – y^p)/(x – y) = s^p$,

for some $r,s$ with $\gcd(r,s) = 1$.

We now proceed to prove some corollary:

Corollary:

$\quad$ There can be no solutions to our system.

Proof:

Because $\gcd(x – y,(x^p – y^p)/(x – y)) = \gcd(x – y,z^p/(x – y) = 1$, there exist some $a,b$ such that:

$ \quad \quad \quad \quad a(x – y) + bz^p/(x – y) = 1$

$ \quad \quad \quad \quad \implies a(x – y)^2 – (x – y) + bz^p = 0$

We now use the quadratic formula to show there can be no solutions to our system by infinitely generating smaller ones.

$ \quad \quad \quad \quad D_{x – y} = 1 – 4abz^p = d^2$, for some d

$ \quad \quad \quad \quad \implies (1 – d)(1 + d) = 4abz^p$

so $d$ is odd, say $2e + 1$ for some $e$.

$ \quad \quad \quad \quad \implies -2e(2e + 2) = 4abz^p$

$ \quad \quad \quad \quad \implies -e(e + 1) = abz^p$

$ \quad \quad \quad \quad \implies e^2 + e + abz^p = 0$

$ \quad \quad \quad \quad \implies D_e = 1 – 4abz^p = f^2$, for some f.

$ \quad \quad \quad \quad \implies (1 – f)(1 + f) = 4abz^p$

so $f$ is odd, say $2g + 1$, for some $g$.

$ \quad \quad \quad \quad \implies -2g(2g + 2) = 4abz^p$

$ \quad \quad \quad \quad \implies -g(g + 1) = abz^p$

$ \quad \quad \quad \quad \implies g^2 + g + abz^p = 0$

We see we keep ending up on the equation:

$ \quad \quad \quad \quad u^2 + u + abz^p = 0$

Since we can infinitely repeat this procedure through calculation of the discriminant $D_u$, we find there is no smallest solution.

We conclude our system cannot have any solutions.