The only groups of order $p^n$ that have a unique subgroup of order $p^{n-1}$ are the cyclic groups of order $p^n$.
This follows because:
Proposition. A group is a union of proper subgroups if and only if it is not cyclic. In particular, a group with a unique maximum proper subgroup is cyclic.
Proof. If $G$ is not cyclic, then for all $x\in G$ we have that $\langle x\rangle$ is a proper subgroup of $G$. Since $$G=\cup_{x\in G}\{x\}\subseteq \cup_{x\in G}\langle x\rangle \subseteq G$$ it follows that $G$ is the union of the proper subgroups $\{\langle x\rangle\mid x\in G\}$.
Conversely, if $G$ is cyclic, $G=\langle x\rangle$, and $G$ is a union of subgroups $G=\cup_{i\in I}H_i$, then $x\in H_i$ for some $i\in I$, and therefore, $G=\langle x\rangle\subseteq H_i\subseteq G$, hence $H_i=G$. So if $G$ is a union of subgroups, at least one of them is not a proper subgroup. $\Box$
So now suppose $G$ has order $p^n$. Subgroups of order $p^{n-1}$ are maximal, and every proper subgroup is contained in a subgroup of order $p^{n-1}$ (this is often proven as a consequence or part of Sylow's First Theorem). So if $G$ has a unique subgroup of order $p^{n-1}$, then the union of all its proper subgroups is the subgroup of order $p^{n-1}$, so $G$ is cyclic. Conversely, if $G$ is cyclic then it has a unique subgroup of any given order, in particular of order $p^{n-1}$.
Now, it is true that if $p^k$ divides the order of $G$, then the number of subgroups of order $p^k$ is congruent to $1$ modulo $p$: this was proven by Frobenius as a generalization of Sylow's Third Theorem, and you can find a proof here. However, in general you do not know that it divides the order $G$ unless $p^k$ is the largest power of $p$ that divides the order of $G$ (in which case, it is part of Sylow's Third Theorem). In fact, it need not divide the order, as the Klein $4$-group (or any group of the form $C_p\times C_p$) shows: the Klein $4$-group has $3$ subgroups of order $2$, and while $3\equiv 1\pmod{2}$, $3$ does not divide $4$. Likewise, the group $C_p\times C_p$ has $p+1$ subgroups of order $p$, which does not divide $p^2$.
The fact that the subgroups of index $p$ are normal follows in any number of ways: it is well known that the normalizer of a proper subgroup of a $p$-group always properly contains the subgroup. Or you can use the also very well known group theory result that if the index of $H$ is the smallest prime that divides the order of $G$, then $H$ is normal.
