I'm in middle of reading a paper (in the appendix, near the end) which is using a trick that I can not follow. The argument follows as such:
The proof is for simplicity of a polynomial, with degree $d = 2^e$. The $e$ will appear later, the rest is just for context.
Let $\nu$ be a valuation on some number field $\mathbb{Q}(b)$ extending $\nu_2$ the $2$ valuation on $\mathbb{Q}$.
If $\nu(x),\nu(y) \ge 0$, then $\nu((x+y)-(x-y)) = \nu(2y) \ge 1$. So if $\nu(x-y) = 0,$ then $\nu(x+y) = 0$. Hence $\nu(x^2 - y^2) = 0$ and thus $\nu(x^{2^k} - y^{2^k}) = 0$ for any $k \ge 0$.
In particular, we have $\nu(x^{2^e} - y^{2^e}) > 0 $ implies $\nu(x-y) > 0$. There are no special properties which I have not mentioned above as far as I can see.
I've been working through those sentences to prove the claims, I've proved up to "in particular". Do we not need the $\nu(x),\nu(y)\ge 0$ assumption? (this is used later, but not checked)
Proofs of stuff I can do:
The first claim, assuming $\nu(x),\nu(y)\ge 0$, we have $\nu(2y) = \nu(2) + \nu(y) \ge 1 + 0 = 1$.
If $\nu(x - y) = 0$, then $0 = \nu(x - y) \ge \min(\nu(x + y), \nu(2y))$. As the second term is greater than $0$, we have that $\nu(x + y) \le 0$. And $(\nu(x+y) \ge \nu(x) + \nu(y) \ge 0$, giving the equality $\nu(x+y) = 0$ So $\nu(x^2 - y^2) = \nu(x+y) + \nu(x-y) = 0$.
Assume that for some $k \ge 1$, we have that $\nu(x^{2^k} - y^{2^k}) = 0$. Then $0 = \nu(x^{2^k} + y^{2^k} - 2y^{2^k}) \ge \min(\nu(x^{2^k} + y^{2^k}), \nu(2y^{2^k}))$. The second term is greater than $0$ due to the factor of $2$, we have that $\nu(x^{2^k} + y^{2^k}) = 0$ with similar reasons to above, and so $\nu((x^{2^k} + y^{2^k})(x^{2^k} - y^{2^k})) = 0 = \nu(x^{2^{k+1}} - y^{2^{k+1}})$. By induction we have that $\nu(x^{2^k} - y^{2^k}) = 0$ for all $k \ge 1$.
Consider $\mathbb{Q}$.
Let $x = \frac{3}{2}$ and $y = \frac{1}{2}$. Then,
$$ \nu(x-y) = v(1) = 0 $$ $$ \nu(x^2 - y^2) = v(2) = 1$$
so $\nu(x^2 - y^2) > 0$, on its own, is insufficient to deduce $\nu(x-y) > 0$.
In your own work, the hypothesis that $\nu(y) \geq 0$ was presumably invoked in order to conclude $\nu(2y) > 0$.