If you only know that your poset is graded, I don't see any easy bounds, unless you have some auxiliary information. In general, the $y$'s could be all elements of rank $k$ (except $x$ itself), for example if your poset looks like this, and you fix $m=1$, $k=3$, $x=11$. Every distinct pair of elements at rank $k=3$ meet at rank $m=1$, namely, at the element "1".
At the other extreme, if you know your poset is the Boolean lattice on a ground set of $n$ elements, the answer is simple. Seeing that $x$ and $y$ are sets of $k$ elements, their meet has rank $m$ iff $|x \cap y|=m$. Given $x$, the number of such $y$'s is $$ \binom{k}{m} \binom{n-k}{k-m} $$ because they are exactly those $k$-element sets that have $m$ elements from $x$, and $k-m$ elements from outside $x$.
However, it sounds like you have some case in between. It may be easier to answer if you can specify your case in more detail.