You can easily count the number of maximal subgroups of $W(k)$, the $k$-fold interated wreath product of $\mathbb{Z}_{2},$ by calculating the index of the Frattini subgroup. You can inductively prove that the number of generators is $k,$ which is clear for $k =1,2.$ To proceed, note that $W(k) = W(k-1) \wr \mathbb{Z}_{2}.$ Factor out the Frattini subgroup of the base group, and by induction, you are left with $E(k-1) \wr \mathbb{Z}_{2}$, where $E(k-1)$ is elementary Abelian of order $2^{k-1}.$ If $x$ is an element of order $2$ outside the new base group, then $[E(k-1),x]$ [E(k-1) \times E(k-1),x]$has order$2^{k-1},$so that the largest elementary Abelian factor group of the original wreath product does have order$2^{k},$as claimed. Hence the group$W(k)$has$2^{k}-1$maximal subgroups, since there is a bijection between maximal subgroups of$W(k)$and maximal subgroups of$W(k)/\Phi(W(k)).$2 revised text to clarify which groups being discussed You can easily count the number of maximal subgroups of$A(k)$, W(k)$, the $k$-fold interated wreath product of $\mathbb{Z}_{2},$ by calculating the index of the Frattini subgroup. You can inductively prove that the number of generators is $k,$ which is clear for $k =1,2.$ To proceed, note that $A(k) W(k) = A(k-1W(k-1) \wr \mathbb{Z}_{2}.$ Factor out the Frattini subgroup of the base group, and by induction, you are left with $E(k-1) \wr \mathbb{Z}_{2}$, where $E(k-1)$ is elementary Abelian of order $2^{k-1}.$ If $x$ is an element of order $2$ outside the new base group, then $[E(k-1),x]$ has order $2^{k-1},$ so that the largest elementary Abelian factor group of the original wreath product does have order $2^{k},$ as claimed. Hence your the group $A = A(k)$ W(k)$has$2^{k}-1$maximal subgroups, since there is a bijection between maximal subgroups of$A$W(k)$ and maximal subgroups of$A/\Phi(A).$of $W(k)/\Phi(W(k)).$
You can easily count the number of maximal subgroups of $A(k)$, the $k$-fold interated wreath product of $\mathbb{Z}_{2},$ by calculating the index of the Frattini subgroup. You can inductively prove that the number of generators is $k,$ which is clear for $k =1,2.$ To proceed, note that $A(k) = A(k-1) \wr \mathbb{Z}_{2}.$ Factor out the Frattini subgroup of the base group, and by induction, you are left with $E(k-1) \wr \mathbb{Z}_{2}$, where $E(k-1)$ is elementary Abelian of order $2^{k-1}.$ If $x$ is an element of order $2$ outside the new base group, then $[E(k-1),x]$ has order $2^{k-1},$ so that the largest elementary Abelian factor group of the original wreath product does have order $2^{k},$ as claimed. Hence your group $A = A(k)$ has $2^{k}-1$ maximal subgroups, since there is a bijection between maximal subgroups of $A$ and maximal subgroups of$A/\Phi(A).$