This is true simply because $\,\Delta = 2\,$ is the determinant of the linear map $\rm\: (x,y)\,\mapsto\, (x\!-\!y,\, x\!+\!y).\:$ More generally, inverting a linear map by Cramer's Rule, i.e. $\rm\color{#0a0}{scaling}$ by the $\rm\color{#c00}{adjugate}$ yields

$$\begin{align} \rm\color{#c00}{\begin{bmatrix}\rm\ \, d &\!\!\! \rm -b \\ \!\!\rm -c & \rm a \end{bmatrix}} \color{#0a0}\times&\, \left\{\, \begin{bmatrix}\rm a & \rm b \\ \rm c & \rm d \end{bmatrix} \begin{bmatrix} \rm x \\ \rm y \end{bmatrix} \,=\, \begin{bmatrix}\rm X \\ \rm Y\end{bmatrix}\, \right\}\\[.2em]
\Longrightarrow\!\!\!\!\!\!\!\!\!\!\!\!\! &\,\qquad\qquad\, \begin{array}\ \rm\Delta\ x\ =\, \ \ \rm \color{#c00}d\ X \color{#c00}{- b}\ Y \\ \rm\Delta\ y\ = \rm \color{#c00}{-c}\ X + \color{#c00}a\ Y \end{array}^{\phantom{|}} ,\ \ \ \ \rm \Delta\ :=\ \color{#c00}{ad-bc} \end{align}\qquad$$

Hence $\rm\ n\ |\ X,Y\ \Rightarrow\ n\ |\ \Delta\:x,\Delta\:y\ \Rightarrow\ n\ |\ gcd(\Delta\, x,\Delta\,y)= \Delta\ gcd(x,y)\,$ by here & here.

In particular, if $\rm\:gcd(x,y) = 1\:$ and $\rm\,\Delta\,$ is prime, we conclude that $\rm\:gcd(X,Y) = 1\:$ or $\rm\:\Delta$.

Your problem is simply the special case $\rm\ a = c = d = 1,\ b = -1\ \Rightarrow\ \Delta = ad\!-\!bc = 2$.

**Remark** $ $ By above we infer $\rm \,n:=\gcd(X,Y)\mid \Delta \gcd(x,y).\,$ Further we have $\rm\ \gcd(x,y)\mid X,Y\,\Rightarrow\, \gcd(x,y)\mid \gcd(X,Y)^{\phantom{|^|}}$ hence

**Theorem** $\rm\,\ (x,y)\overset{A}\mapsto (X,Y)\,$ linear $\,\Rightarrow\, \bbox[7px,border:1px solid #c00]{\rm\gcd(x,y)\mid \gcd(X,Y)\mid det(A) \gcd(x,y)}$

Worth emphasis: this has a nice arithmetical interpretation in Gaussian integer arithmetic, where the linear map is simply multiplication by $\rm 1 + \it i.\:$ See my other answer here for details.

The above can also be viewed more geometrically in terms of lattices.