Can you replace $x$ with $x^2$ for any Maclaurin series?

Yes, because you are evaluating the function at the point $x^2$ instead of just $x$. All you are really doing is chaning the domain of the function, so that if the original radius of convergence was $r$, the new radius of convergence is $\sqrt r$.

If $$f(x)=\sum_{n\ge0}a_nx^n \text{ for }\vert x \vert < R,$$ then $$f(x^k)=\sum_{n\ge0}a_nx^{nk} \text{ for }\vert x^k \vert < R\text{, i.e.} \vert x \vert <R^{1/k}$$

$${{{{{{\mathbf{\text{Yes}.}}}}}}}$$