I was wondering if the following statement is correct:
Let $P(x,y)$ be a real polynomial, and $\gcd(r,s)=1$. If $y-x^s$ divides $P(x^r,y)$, then $y^r-x^{rs}$ divides $P(x^r,y)$ (both in the real field).
My idea is that if $y-x^s$ divides $P(x^r,y)$ and since $\gcd(r,s)=1$, the polynomial $P(x^r,y)$ must also have all corresponding ``conjugate factors" to cancel out the terms of $P(x^r,y)$ with order of $x$ not divisible by $r$.
The statement is correct. We show that the conjugate factors also divide $P(x^r,y)$ as you suggested.
Let $Q(x,y)=P(x^r,y)$. For a $r$-th root of unity $\zeta\in\Bbb C$ consider the homomorphism $\psi:\Bbb C[x,y]\to\Bbb C[x,y]$ induced by $x\mapsto \zeta x,y\mapsto y$. Note that we have $\psi(Q)=Q$. Thus $\psi(y-x^s)=y-\zeta^sx^s\mid\psi(Q)=Q$ in $\Bbb C[x,y]$. As $(r,s)=1$ the map $\zeta\mapsto \zeta^s$ is a permutation of the $r$-th roots of unity. We conclude that $y-\zeta x^s\mid Q$ for all $\zeta$ such that $\zeta^r=1$. This implies $$y^r-x^{rs}=\prod_{\zeta^r=1}(y-\zeta x^s)\mid Q(x,y)=P(x^r,y)$$ in $\Bbb C[x,y]$ and therefore also in $\Bbb R[x,y]$.