Suppose $G$ is a finite non-abelian p-group of nilpotent class $c$. Is there a subgroup $H$ of nilpotent class $c$ and size $p^{c+1}$?
If this is not true, is it possible to add some additional assumptions to satisfy the sentence?
The answer to the question is no for $p=2$. The $2$-groups of maximal class have been classified. There are only $3$ isomorphism types and they all have a normal cyclic subgroup of order $2^c$. But there are $2$-groups of exponent $4$ with arbitrary large class.
For a general method of constructing counterexamples, let $P$ be any $p$-group of maximal class $c >1$. Then $\Phi(P)=[P,P]$ and $|P/\Phi(P)|=p^2$.
Let $Q = C_{p^2}^2$, and let $G$ be a subdirect product of $P \times Q$ that contains $\Phi(P) \times \Phi(Q)$ such that $|G:(\Phi(P) \times \Phi(Q))|=p^2$. So $|G| = p^{c+1+2}$. Since $\Phi(P) = [G,G] \le \Phi(G)$ and $G^p\Phi(P) = \Phi(Q)$, we have $\Phi(G) = \Phi(P) \times \Phi(Q)$.
So any maximal subgroup of $G$ maps onto a proper subgroup of $G/(\Phi(P) \times \Phi(Q))$ and hence onto a proper subgroup of $P$, and hence has class less than $c$. So $G$ cannot contain a subgroup having maximal class.