Understanding the free $R$-algebras.

Question

Let $R$ be a commutative ring, $R\left< x,y \right>$ the free $R$-algebra on indeterminates $x$ and $y$. If $z_i=xy^i$, I want to show that the $R$-subalgebra generated by the elements $z_i$ is itself a free $R$-algebra. However, I'm not really sure what it is I'm supposed to be showing. My professor told me that the easiest way to go about doing this is show the map $$R\{x,xy,\cdots,xy^n\} \to R\{x,y\}, $$ is an injection. But I don't understand why this immediately gives us that $R\{x,xy,\cdots, xy^n\}$ is free. It seems to me like I must have misheard him. What he seems to be suggesting is that we may take any collection of monomials in $R\{x,y\}$, then they must generate a free $R$-algebra, since any inclusion map $R\{\cdot \} \to R\{x,y\}$ is going to be an injection, since only $0$ can map to $0$. If this were the case though, why would he not just ask us to prove this statement? Moreover, I can't find anything online to back this claim, so there's clearly something I'm not understanding.