How does this series scale ? (A fractional Touchard polynomial)

By the Cauchy-Schwarz inequality, for any $x\geq 0$ we have

$$ \sum_{k\geq 1}\frac{x^k}{k!}\sqrt{k}\leq \sqrt{\sum_{k\geq 1}\frac{x^k}{k!}\sum_{k\geq 1}\frac{x^k}{(k-1)!}}=\sqrt{x e^x(e^x-1)}\leq \sqrt{x}\,e^{x} \tag{1}$$ which is enough to settle the question.

An alternative approach is to directly square $f(x)=\sum_{k\geq 0}\frac{x^k}{k!}\sqrt{k}$:

$$ f(x)^2 = \sum_{n\geq 0}\frac{x^n}{n!}\sum_{k=0}^{n}\binom{n}{k}\sqrt{k(n-k)}\leq \sum_{n\geq 0}\frac{n(2x)^n}{2n!}=x e^{2x}.\tag{2} $$

An interesting follow-up might be to compute $$ \lim_{x\to +\infty}\frac{1}{\sqrt{x}e^x}\sum_{k\geq 0}\frac{x^k}{k!}\sqrt{k}. $$ By the previous estimates this limit is $\leq 1$, but since over $z\in[0,1]$ we have $\sqrt{z(1-z)}\geq 2z(1-z)$, for any $x\geq\tfrac{1}{2}$ we also have $$ f(x)^2 \geq \sum_{n\geq 0}\frac{x^n}{n!}\sum_{k=0}^{n}\binom{n}{k}2k\left(1-\frac{k}{n}\right)=\sum_{n\geq 0}\frac{(n-1)(2x)^n}{2n!}=\left(x-\tfrac{1}{2}\right)e^{2x} \tag{3}$$ from which it follows that the previous limit equals one and $f(x)\sim\sqrt{x} \,e^x$ as $x\to +\infty$.

What about deriving more terms of the asymptotic expansion? First, we may notice that $g(x)=\sqrt{x}\,e^x$ is a solution of the differential equation $g'(x)=\left(1+\frac{1}{2x}\right)g(x)$.
On the other hand

$$ f'(x)-\left(1+\tfrac{1}{2x}\right)f(x)=\sum_{k\geq 0}\frac{x^k}{k!}\cdot\frac{1}{2\sqrt{k+1}(\sqrt{k}+\sqrt{k+1})^2} \geq \frac{1}{2}$$ and the asymptotic behaviour of the middle term is $\frac{e^x}{2x^{3/2}}$ by the same squaring approach.
By solving the differential equation $f'(x)-\left(1+\tfrac{1}{2x}\right)f(x)=\frac{e^x}{2x^{3/2}}$ we get an extra term of the asymptotic expansion of $f(x)$. The same can be achieved by noticing that for any $s>1$ we have $$ \left(\mathcal{L}f\right)(s)=\int_{0}^{+\infty}f(x)e^{-sx}\,dx = \sum_{k\geq 1}\frac{\sqrt{k}}{s^{k+1}}\tag{4} $$ and by studying the Laurent series of $(\mathcal{L} f)(z^2+1)$ in a neighbourhod of $z=0$.
For instance, since $$ \sum_{k\geq 1}\frac{\sqrt{k}}{s^{k+1}}=\frac{\sqrt{\pi}}{2(s-1)^{3/2}}-\frac{\sqrt{\pi}}{8(s-1)^{1/2}}+\zeta\left(-\tfrac{1}{2}\right)+O\left(\sqrt{s-1}\right)\tag{5} $$ in a right neighbourhood of $s=1$, by applying $\mathcal{L}^{-1}$ to both sides of $(5)$ we get $$ f(x)=\sum_{k\geq 1}\frac{x^k}{k!}\sqrt{k}\stackrel{x\to +\infty}{\sim}\color{blue}{\left(\sqrt{x}-\frac{1}{8\sqrt{x}}\right)e^x}.\tag{6}$$

As a further addendum, I would like to give a more explicit touch to Marco's approach. For any $m\geq 0$ we have $$\sum_{k\geq 0}\frac{x^{k+1}}{(k+1)!}(k+1)^m = T_m(x) e^x = e^x\sum_{k=0}^{m}{m\brace k}x^k \tag{7}$$ hence we get that: $$ f(x)=\sum_{k\geq 1}\frac{x^k}{k!}\sqrt{k} = e^x\sum_{h\geq 0}T_h(x)\sum_{u\geq 0}\binom{h+u}{h}\binom{1/2}{h+u}(-1)^u \tag{8}$$ where $\sum_{h\geq 0}x^{h-\tau}\binom{h+u}{u}\binom{1/2}{h+u} $ behaves like $\frac{C}{(1+x)^{u+\tau-\frac{1}{2}}}$ as $x\to +\infty$. It follows that the first two terms of the asymptotic expansion of $f(x)$ can be recovered from the previous formula, restricted to the range $\tau\in\{0,1\}$ and $u\in\{0,1\}$, i.e $$ \frac{f(x)}{e^x}\sim\sum_{h\geq 0}\left(x^h+\binom{h}{2}x^{h-1}\right)\cdot\left(\binom{1/2}{h}-\binom{1/2}{h+1}(h+1)\right) $$ returning $(6)$ as expected.

By combining both approaches, it is clear the full asymptotic expansion of $\frac{f(x)}{e^x}$ is given by a Touchard polynomial with fractional order, $T_{1/2}(x)$. I am not an expert on Stirling numbers of the second kind with fractional order, but I guess this peculiar application of fractional calculus has already been studied in depth. For instance, we may notice that from the classical approximation $\frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi n}}$ it follows that $$ f(x)\approx\sum_{k\geq 1}\frac{x^k\sqrt{\pi}}{(k-1)!4^k}\binom{2k}{k}=\frac{x}{2}e^{x/2}\sqrt{\pi}\left[I_0\left(\frac{x}{2}\right)+I_1\left(\frac{x}{2}\right)\right]$$ where the asymptotic behaviour of modified Bessel functions of the first kind is pretty well-known, as a consequence of Laplace's method. By combining $\sqrt{k}\approx\frac{\sqrt{\pi}k}{4^k}\binom{2k}{k}$ with the Cauchy-Schwarz inequality, we get: $$ f(x)\leq\sqrt{\frac{x^{3/2}}{2}e^{3x/2}\sqrt{\pi}\left[I_0\left(\frac{x}{2}\right)+I_1\left(\frac{x}{2}\right)\right]\text{Erf}(\sqrt{x})} $$ and $$\forall x\geq 1,\qquad \sum_{k\geq 1}\frac{x^k}{k!}\sqrt{k}\leq \color{blue}{e^x\sqrt{x-\tfrac{1}{4}}},\tag{9}$$ a remarkable inequality capturing the first $2$ terms of the asymptotic expansion. I have just found a simpler proof of $(9)$: by considering the Taylor series of $\sqrt{x(1-x)}$ at $x=\frac{1}{2}$, we have that $$\begin{eqnarray*} f(x)^2&=&\sum_{n\geq 0}\frac{x^n}{n!}\sum_{k=0}^{n}\binom{n}{k}\sqrt{k(n-k)}\\&\leq& \sum_{n\geq 0}\frac{x^n}{n!}\sum_{k=0}^{n}\binom{n}{k}n\left(\tfrac{1}{2}-\left(\tfrac{k}{n}-\tfrac{1}{2}\right)^2\right)\\&=&\sum_{n\geq 0}\frac{(2x)^n}{n!}\cdot\frac{2n-1}{4}=\left(x-\tfrac{1}{4}\right)e^{2x}.\tag{10}\end{eqnarray*}$$

By the generalized binomial theorem we have $$\sum_{k=1}^{\infty}\frac{x^{k}}{k!}\sqrt{k}=\sum_{k=0}^{\infty}\frac{x^{k+1}}{\left(k+1\right)!}\sqrt{k+1}=\sum_{k=0}^{\infty}\frac{x^{k+1}}{\left(k+1\right)!}\sum_{m=0}^{\infty}\dbinom{1/2}{m}k^{m}$$ $$=\sum_{m=0}^{\infty}\dbinom{1/2}{m}\sum_{k=0}^{\infty}\frac{x^{k+1}}{\left(k+1\right)!}k^{m}$$ and note that $$\sum_{k=0}^{\infty}\frac{x^{k+1}}{\left(k+1\right)!}k^{m}=g_{m}\left(x\right)$$ where $g_{m}\left(x\right)$ is of the form $$g_{m}\left(x\right)=\sum_{\ell=0}^{m}a_{\ell}e^{x}x^{\ell}$$ where $a_{m}=1$. So $$\sum_{k=1}^{\infty}\frac{x^{k}}{k!}\sqrt{k}=e^{x}\sum_{m=0}^{\infty}\dbinom{1/2}{m}x^{m}+e^{x}\sum_{m=0}^{\infty}\dbinom{1/2}{m}\sum_{\ell=0}^{m-1}a_{\ell}e^{x}x^{\ell}$$ $$=e^{x}\sqrt{x+1}+e^{x}\sum_{m=0}^{\infty}\dbinom{1/2}{m}\sum_{\ell=0}^{m-1}a_{\ell}x^{\ell}$$ then