Consider the ring $M:=\mathbb F_p[[x,y]]$ and subring $N:=\mathbb F_p[[x^{p^m},y^{p^n}]]$, where $\mathbb F_p$ is the field of $p$-elements and $\gcd(m,n)=1$. Then $M$ is a finitely generated module over $N$ with the generating set $G:=\{x^iy^j:~0 \leq i \leq p^m-1,~0 \leq j \leq p^n-1\}.$
Want to show $G$ is indeed a basis for $M$. Or in other words, the rank of $M$ is $p^{m+n}$.
Method1:
Since $G$ is a generator of $M$ over $N$, any element $m(x,y) \in M$ can be written as $m(x,y)=g_1n_1(x^{p^m},y^{p^n})+g_2n_2(x^{p^m},y^{p^n})+\cdots+(\text{upto finite combination})$, where $g_1, g_2, \cdots \in G$. But still, we have to show the minimality $G$. If we remove just one element $x^i y^j$ from $G$, then we have to show there is at least one element $m(x,y \in M$ which cannot be written as a finite linear combination of $G \setminus \{x^iy^j\}$ and $N$.
If I consider the particular case for $m=1, n=2$: then we have
$G=\{1, x, y, xy, xy^2, xy^3\}$, $N=\mathbb F_2[[x^2, y^4]]=\left\langle \mathbb F_2, x^2, y^4 \right\rangle$ and $M=\mathbb F_2[[x,y]]$.
Let us remove $y$ from the generating set $G$ and consider the element $m(x,y)=x^5+y^6 \in \mathbb F_2[[x,y]]$. We claim that $m(x,y)$ cannot be written as a finite linear combination of elements of the sets $G':=G \setminus \{x\}=\{1, y, xy, xy^2, xy^3\}$ and $N$.
If possible, let $x^5+y^{6}= g_1n_1+g_2n_2,~g_i \in G',~n_i \in N$, then we must have $g_1n_1=x^5$ and $g_2n_2=y^{6}$.
Clearly, $g_1n_1=x^5$ cannot have a solution in $G'$, this only can happen if $g_1=x, n_1=(x^2)^2.$. But $x \notin G'$. Thus, $G'$ cannot generate $M$. In other words, $G$ is the minimal generating set. So a basis for $M$, in this case.
How to follow the general case ?
With the help of the comments
by Mariano Suárez-Álvarez, I am writing the answer:
For more generality, given a field $k$, consider the corresponding ring $M:=k[[x,y]]$ of power series and its subring $N:=k[[x^m,y^n]],~m,n \in \mathbb N$. Then, $M$ is a $N$-module where the action of $N$ on $M$ is just multiplication of power series. Besides that, $M$ is finitely generated by the set $\Theta:=\{x^iy^j:~0 \leq i \leq m, 0\leq j \leq n\}$. We want to show the set $\Theta$ is indeed a basis for $M$. For example, consider the linear combination \begin{align} &\sum_{l=1}^{mn} c_i \cdot \left( \sum_{i=0}^{m-1} \sum_{j=0}^{n-1}x^iy^j\right)=0,~(\text{where}~c_i:=c_i(a,x^m,y^n) \in N,~a \in \mathbb F_p) \\& \Rightarrow c_1 \cdot 1+c_2 \cdot x+c_3 \cdot y+c_4 \cdot xy+\cdots+c_{mn}x^{m-1}y^{m-1}=0.\end{align} The LHS $c_1 \cdot 1+c_2 \cdot x+c_3 \cdot y+c_4 \cdot xy+\cdots+c_{mn}x^{m-1}y^{m-1}$ vanishes if and only if $c_i=p=0 \in \mathbb F_p$. Therefore, the generating set $\Theta$ is linearly independent and hence a basis for $M$ over $N$. Note that $\# \Theta=mn$ and thus the free-rank of $M$ is $mn$.
Finally, the answer to the given question follows trivially by replacing $(m,n) \to (p^m,p^n)$.