coefficient of operator for $B_{n,k}^{x^2}(x)$ We start with the following:
$$
(x+z)^2 - x^2 = \sum_{n \geq 1} \frac{z^n}{n!} \frac{d^n}{dx^n}[x^2]
$$
$$
(x+z)^2 - x^2 = z(2x+z)
$$
$$
z^k(2x+z)^k = \sum_{n \geq k} Y^{\Delta}(n,k,x)z^n
$$
Where
$$
Y^{\Delta}(x) = \frac{k!}{n!}B_{n,k}^{x^2}(x)
$$
Therefore:
$$
(2x)^k [z^{n-k}]\left(1+\frac{z}{2x}\right)^k
$$
$$
= (2x)^k [z^{n-k}] \sum_{j=0}^k {k \choose j} \left(\frac{z}{2x}\right)^k
$$
$$
= (2x)^{2k-n} {n-1 \choose n-k}
$$
Is this correct? I feel like this is very wrong.
 A: Only the last calculation is somewhat wrong (see Hints below). Everything else is ok.

Let's start from 
  \begin{align*}
\left[(x+z)^2-x^2\right]^k=z^k(2x+z)^k = \sum_{n \geq k} Y^{\Delta}(n,k,x)z^n\tag{1}
\end{align*}
  Then we know from this paper by V. Kruchinin, that
  \begin{align*}
Y^{\Delta}(n,k,x) = \frac{k!}{n!}B_{n,k}^{x^2}(x)
\end{align*}

(Hint: $Y^{\Delta}$ has three arguments)

We obtain
  \begin{align*}
\frac{k!}{n!}B_{n,k}^{x^2}(x)&=[z^n]\sum_{n \geq k} Y^{\Delta}(n,k,x)z^n\\
&=[z^n]z^k(2x+z)^k\\
&=(2x)^k[z^{n-k}]\left(1+\frac{z}{2x}\right)^k\\
&=(2x)^k[z^{n-k}]\sum_{j=0}^k\binom{k}{j}\left(\frac{z}{2x}\right)^j\tag{2}\\
&=(2x)^k\binom{k}{n-k}\left(\frac{1}{2x}\right)^{n-k}\\
&=(2x)^{2k-n}\binom{k}{n-k}\tag{3}
\end{align*}

(Hint: The exponent of $\frac{z}{2x}$ in (2) is $j$, not $k$)
We conclude from (3)

The Bellpolynomials $B_{n,k}^{x^2}(x)$ are 
  \begin{align*}
B_{n,k}^{x^2}(x)=\frac{n!}{k!}\binom{k}{n-k}(2x)^{2k-n}\tag{4}
\end{align*}

We can generalise the problem from $x^2$ to $x^N$ with $N>0$ a positive integer. A nice by-product is, when specializing the result to $N=2$ we will derive a binomial identity.

Generalisation:
We consider the $k$-th power of $Y(x+z)=(x+z)^N-x^N$ and obtain
  \begin{align*}
[Y(x,z)]^k&=[(x+z)^N-x^N]^k\\
&=x^{Nk}\left[\left(1+\frac{z}{x}\right)^N-1\right]^k\\
&=x^{Nk}\sum_{j=0}^k\binom{k}{j}\left(1+\frac{z}{x}\right)^{Nj}(-1)^{k-j}\\
\end{align*}
Similarly to (1) we observe
  \begin{align*}
[(x+z)^N-x^N]^k=x^{Nk}\sum_{j=0}^k\binom{k}{j}\left(1+\frac{z}{x}\right)^{Nj}(-1)^{k-j}=\sum_{n \geq k} Y^{\Delta}(n,k,x)z^n
\end{align*}

Again, we know from this paper by V. Kruchinin, that
\begin{align*}
Y^{\Delta}(n,k,x) = \frac{k!}{n!}B_{n,k}^{x^N}(x)
\end{align*}
It follows
\begin{align*}
\frac{k!}{n!}B_{n,k}^{x^N}(x)&=[z^n]\sum_{n \geq k} Y^{\Delta}(n,k,x)z^n\\
&=[z^n]x^{Nk}\sum_{j=0}^k\binom{k}{j}\left(1+\frac{z}{x}\right)^{Nj}(-1)^{k-j}\\
&=x^{Nk}\sum_{j=0}^k\binom{k}{j}[z^n]\left(1+\frac{z}{x}\right)^{Nj}(-1)^{k-j}\\
&=x^{Nk}\sum_{j=0}^k\binom{k}{j}\binom{Nj}{n}x^{-n}(-1)^{k-j}\\
&=x^{Nk-n}\sum_{j=0}^k\binom{k}{j}\binom{Nj}{n}(-1)^{k-j}
\end{align*}

We conclude, the Bellpolynomials $B_{n,k}^{x^N}(x)$ are 
  $$B_{n,k}^{x^N}(x)=x^{Nk-n}\frac{n!}{k!}\sum_{j=0}^k\binom{k}{j}\binom{Nj}{n}(-1)^{k-j}$$
The special case $N=2$ results in
  $$B_{n,k}^{x^2}(x)=x^{2k-n}\frac{n!}{k!}\sum_{j=0}^k\binom{k}{j}\binom{2j}{n}(-1)^{k-j}$$

Comparing this result with (4) above we observe

the following binomial identity is valid for $n\geq 1$
\begin{align*}
\sum_{j=\lfloor\frac{n}{2}\rfloor}^k\binom{k}{j}\binom{2j}{n}(-1)^{k-j}=2^{2k-n}\binom{k}{2k-n}\qquad\quad 1\leq k \leq n
\end{align*}

