As the ring of Gaussian integers is a UFD, this means that $a^2 \mid b^2$ leads to $a\mid b$.
Is there any subring of the ring of Gaussian integers with infinitely many elements such that $a^2\mid b^2$ does not lead to $a\mid b$?
If so, then is there a set $X$ of elements such that for $\forall x, y \in X$ where $x \neq y$, $x^2 \mid y^2$ but $x \not\mid y$?
For example, let $R$ be the subring of Gaussian integers of the form $a + b i$ where $b$ is even. Take $a = 2 + 2i$ and $b = 2 + 6 i$. Then $a^2 | b^2$ in $R$ because $b^2/a^2 = 3 + 4 i \in R$, but $a \not\mid b$ since $b/a = 2 + i \notin R$.
If both $x^2 | y^2$ and $y^2 | x^2$, $x^2/y^2$ must be a unit, but there are only four units $\pm 1, \pm i$ in the Gaussian integers. If $i$ or $-i$ is in $R$, then $R$ is all the Gaussian integers. So $x^2/y^2 = \pm 1$. It can't be $+1$ else $x/y = -1$, but a subring that contains $1$ contains $-1$. So it must be $-1$, and $x/y = \pm i$. Indeed, with $R$ as above we can take $X = \{2,2i\}$ where $(2i)^2 | 2^2$ and $2^2 | (2i)^2$ because $-1 \in R$, but $2 \not\mid 2i$ and $2i \not\mid 2$ because $i$ and $1/i = -i$ are not in $R$.