In the book Catalan's conjecture by R. Schoof, Lemma 3.2 on page 14 is the following:
Let $q\geq3$ be prime and let $x$, $y$ be nonzero integers satisfying $x^2-y^q=1$. Then we have $x\equiv0\pmod{q}$.
The proof is about a page long, so I won't reproduce it in full, but it can be found here for example. He shows that $y+1=u^2$ for some nonzero $u\in\Bbb{Z}$, and that $$x+y^{(q-1)/2}\sqrt{y}=\pm(u+\sqrt{y})^m,$$ for some $m\in\Bbb{Z}$. Then I quote (emphasis mine):
Since the elements $1,\sqrt{y}$ for m a $\Bbb{Z}$-basis for the additive group of $\Bbb{Z}[\sqrt{y}]$, it follows that $mu^{m-1}\equiv0\pmod{y}$. This implies that $y$ divides $mu^{m-1}$. Since $y+1=u^2$, Exercise 2.1 implies that $y$ is even and $u$ is odd.
All is good and obvious, until the emphasized sentence. I do not see how it follows. For completeness; Exercise 2.1 is the following very basic exercise:
Let $x\in\Bbb{Z}$. Show that $x^2$ is congruent to $0$ or $1\pmod{4}$. Shows that $x^2+2$ cannot be a square modulo $4$.
So my question is: How does it follow from Exercise 2.1 that $y$ is even and $u$ is odd?