Find a number $x$ such that $\sum_{n=1}^\infty\frac{n^x}{2^n n!} = \frac{1539}{64}e^{1/2}$

One may note that $$\left(z\frac{d}{dz}\right)^a\;\underbrace{\sum_{n=1}^{\infty}\frac{z^n}{n!}}_{e^{z}-1}=\sum_{n=1}^{\infty}\frac{n^a z^n}{n!}.$$ Now indeed setting $a=6$, calculating the derivatives of $e^z-1$ and setting $z=\frac12$ in the final expression, we get $\frac{1539}{64}\sqrt{e}$.

According to Wikipedia, $$\sum_{n = 0}^\infty \frac{n^k z^n}{n!} = e^z T_k(z),$$ where $T_k(z)$ is the $k^\text{th}$ Touchard polynomial. Using $z = \tfrac{1}{2}$ and $k = 6$, we find $$\sum_{n = 0}^\infty \frac{n^6}{2^n n!} = e^{1/2} T_6(1/2) = \frac{1539}{64}e^{1/2}.$$

(Hopefully someone can find a more satisfying solution.)