A heap of size $n$ has atmost $\left \lceil \frac{n}{2^{h+1}} \right \rceil$ nodes with height $h.$
Proof by induction:
Suppose it is true for height $\left ( h-1 \right )$
Let $N_{h}$ be the number of nodes at height $h$ in the $n$ node tree $T.$
Now, the tree $T'$ formed by removing the leaves of $T.$
It has $n'=n-\left \lceil \frac{n}{2} \right \rceil=\left \lfloor \frac{n}{2} \right \rfloor$ nodes.
Let $N'_{h-1}$ denote the number of node at height $\left ( h-1 \right )$ in tree $T'.$
By induction, we have $N_{h}=N'_{h-1}=\left \lceil \frac{n'}{2^{h}} \right \rceil=\left \lceil \frac{n/2}{2^{h}} \right \rceil=\left \lceil \frac{n}{2^{h+1}} \right \rceil$
https://www2.cs.sfu.ca/CourseCentral/307/petra/2009/SLN_2.pdf