Lemma: Let $a,b,x, y \in \Bbb{Z}$. Then $ax^2 = z= by^2 \implies z = \pm \operatorname{lcm}(a,b) \gcd(x,y)^2$.
Proof. If $a,b = \pm 1$ then $z = \pm \operatorname{lcm}(\pm 1, \pm 1) \gcd(\pm x, \pm y)^2 = \pm x^2 = \pm y^2$.
Now assume that it's true for $ab = \pm$ some composition of the first $n\geq 0$ primes. And introduce the next prime $p$. Let $a = p^k u$, $\ b = p^r v$ where $p \nmid u,v$ for some $k, r \geq 0$.
Then $z = p^ku x^2 = p^r v y^2$. Without loss of generality assume that $r \geq k$. Then $\dfrac{z}{p^k} = u x^2 = vp^{r-k} y^2$.
By induction, we have that $\dfrac{z}{p^k} = \pm \operatorname{lcm}(a, vp^{r - k}) \gcd(x,y)^2$. So that $z = \pm \operatorname{lcm}(u, vp^{r - k}) p^k \gcd(x,y)^2 = \pm \operatorname{lcm}(up^k, vp^r) \gcd (x,y)^2$.
Can you verify this proof for me?
If it's true, then that helps prove that $U_a = \{ a x^2 - 1 : x \in \Bbb{Z}\}$ forms a basis for a topology on $\Bbb{Z}$.
Your inductive step doesn't work as written: if you set $z' = z/p^k, a' = u, b' = vp^{r-k}$, then you do indeed have $z' = a'x^2 = b'y^2$, but you don't necessarily know that $a'b'$ doesn't contain a factor of $p$ (because you might have $r > k$).
But your proof can probably be easily modified to show that, in fact, you must have $r=k$.