Bounding the order of a group by its nilpotentizer Let $G$ be a finite non-nilpotent group. We put $nil_G(x)=\{y\in G\mid \langle x,y \rangle \text{ is nilpotent}\}$, called the nilpotentizer of $x$. Note that $nil_G(x)$ may not be a subgroup of $G$, for example in $S_4$, $|nil_{S_4}((12)(34))|=16$. Suppose $|G\setminus nil_G(x)|=n$ for some $x\in G$, then can we find an upper bound for $|G|$ depending on $n$? (The motivation of this question come from the fact that if $|G\setminus C_G(x)|=m$, then $|G|\leq 3m/2$.)Thanks for any help.

This is not a complete answer to the question. I find it hard to believe that for any fixed $n>1$ there could be groups $G$ of arbitrarily large order and $x \in G$ with $|G \setminus {\rm nil}_G(x)| = n$, so I am guessing that the answer is yes there is a bound, but I don't know how to go about proving it.
Here I am just describing examples that show that there can be no such linear bound. The idea is to construct groups with a similar shape to $S_4$ but with the prime $2$ replaced by an arbitrary prime $p$, to construct groups $G$ and $x \in G$ with $|G| > |{\rm nil}_G(x)| > (p-1)|G|/p$.
Let $p$ be any prime and $H$ a Frobenius group with kernel $K$ and complement of order $p$. Now let $G$ be any group with a elementary abelian normal $p$-subgroup $N$ with $G/N \cong H$ such that $N$ is not central in the inverse image of $K$ in $G$, and let $x \in N$ such that $x$ is not centralized by $K$. So, there exists $y \in G$ such that $\langle x,y \rangle$ is not nilpotent, and hence ${\rm nil}_G(x) \ne G$. For example we could take $G$ to be the wreath product of a cyclic group of order $p$ with $H$.
Then, since all element of $H \setminus K$ have order $p$, an element of $G$ lies in ${\rm nil}_G(x)$ whenever its image in $H$ does not lie in $K \setminus \{1\}$, and the proportion of such elements is 
$$\frac{|K|(p-1)-1}{|K|p} > \frac{p-1}{p}.$$
