Upper and Lower Bounds
Note that
$$
e^{-x}\sum_{k=0}^\infty\frac{x^k}{k!}=1\tag{1}
$$
and that
$$
e^{-x}\sum_{k=0}^\infty(k+1)\frac{x^k}{k!}=x+1\tag{2}
$$
Since $\sqrt{x}$ is concave, Jensen's Inequality gives
$$
e^{-x}\sum_{k=0}^\infty\sqrt{k+1}\frac{x^k}{k!}\le\sqrt{x+1}\tag{3}
$$
Also,
$$
e^{-x}\sum_{k=0}^\infty\frac1{k+1}\frac{x^k}{k!}=\frac{1-e^{-x}}{x}\tag{4}
$$
Since $1/\sqrt{x}$ is convex, Jensen's Inequality gives
$$
\begin{align}
e^{-x}\sum_{k=0}^\infty\sqrt{k+1}\frac{x^k}{k!}
&\ge\sqrt{\frac{x}{1-e^{-x}}}\\
&\ge\sqrt{x}\tag{5}
\end{align}
$$
Therefore, we get the bounds
$$
\sqrt{x}\le e^{-x}\sum_{k=0}^\infty\sqrt{k+1}\frac{x^k}{k!}\le\sqrt{x+1}\tag{6}
$$
Asymptotic Expansion
Using Stirling's Expansion and the Binomial Theorem, we get
$$
\begin{align}
\frac1{4^n}\binom{2n}{n}
&=\frac1{\sqrt{\pi n}} \left(1-\frac1{8n}+\frac1{128n^2}+\frac5{1024n^3}-\frac{21}{32768n^4}+\dots\right)\\
&=\frac1{\sqrt{\pi(n+1)}} \left(1+\frac3{8n}-\frac{23}{128n^2}+\frac{89}{1024n^3}-\frac{1509}{32768n^4}+\dots\right)\tag{7}
\end{align}
$$
and therefore,
$$
\begin{align}
\frac{\sqrt{n+1}}{n!}
&=\frac{4^n}{\sqrt{\pi}}\frac{n!}{(2n)!}\left(1+\frac3{8n}-\frac{23}{128n^2}+\frac{89}{1024n^3}-\frac{1509}{32768n^4}+\dots\right)\\
&=\frac{2^n}{\sqrt{\pi}}\frac1{(2n-1)!!}\left(1+\frac3{8n}-\frac{23}{128n^2}+\frac{89}{1024n^3}-\frac{1509}{32768n^4}+\dots\right)\\
&=\frac{2^n}{\sqrt{\pi}}\small\left(\frac1{(2n{-}1)!!}+\frac{3/4}{(2n{+}1)!!}+\frac{1/32}{(2n{+}3)!!}+\frac{9/128}{(2n{+}5)!!}+\frac{491/2048}{(2n{+}7)!!}+\dots\right)\tag{8}
\end{align}
$$
Note that
$$
\begin{align}
\int_x^\infty e^{-t^2/2}\,\mathrm{d}t
&=\frac1x\int_x^\infty\frac{x}{t}e^{-t^2/2}\,\mathrm{d}t^2/2\\
&\le\frac1x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t^2/2\\
&=\frac1xe^{-x^2/2}\tag{9}
\end{align}
$$
therefore, since both the following sum and integral satisfy $f'=1+xf$ and agree at $x=0$,
$$
\begin{align}
\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!!}
&=e^{x^2/2}\int_0^xe^{-t^2/2}\,\mathrm{d}t\\
&=\sqrt{\frac\pi2}\ e^{x^2/2}+O\left(\frac1x\right)\\
\frac1{\sqrt{2x}}\sum_{k=0}^\infty\frac{(2x)^{k+1}}{(2k+1)!!}
&=\sqrt{\frac\pi2}e^x+O\left(\frac1{\sqrt{x}}\right)\\
e^{-x}\sum_{k=0}^\infty\frac{(2x)^{k+1}}{(2k+1)!!}
&=\sqrt{\pi x}+O\left(e^{-x}\right)\tag{10}
\end{align}
$$
Multiplying $(8)$ by $e^{-x}x^n$, summing, and applying $(10)$ yields the asymptotic expansion that Raymond Manzoni got:
$$
\begin{align}
e^{-x}\sum_{n=1}^\infty\frac{\sqrt{n+1}}{n!}x^n
&=\sqrt{x}\small\left(1+\frac3{8x}+\frac1{128x^2}+\frac9{1024x^3}+\frac{491}{32768x^4}+O\left(\frac1{x^5}\right)\right)\tag{11}
\end{align}
$$
If you do not have any further assumption on the values $X$ can take (e.g., is it lower bounded a.s.?), then you cannot get any meaningful lower bound. For any $\varepsilon\in[0,1]$ (and wlog the case $\mu=0$), consider the random variable defined by
$$
X = \begin{cases}
-x\frac{1-\varepsilon}{\varepsilon} & \text{ w.p. } \varepsilon \\
x & \text{ w.p. } 1-\varepsilon
\end{cases}
$$
where $x = \sigma\sqrt{\frac{\varepsilon}{1-\varepsilon}}$.
You have $\mathbb{E} X = 0 = \mu$, and $\operatorname{Var} X = \sigma^2$; yet $\mathbb{P}\{ X < \mu\} = \varepsilon$ can be arbitrarily small.
Assuming $X \geq 0$ a.s. (as suggested in a comment below). Even then, one cannot get a non-trivial bound. Namely,
Fix any $\mu> 0$, $\sigma^2\geq 0$. For any $a\in[0,\mu)$, there exists a random variable $X\in L^2$ such that $X\geq 0$ a.s., satisfying $$\mathbb{P}\{ X \leq a\} = 0.$$
Note that up to renormalization by $\mu$ (of the standard deviation and $a$), we can wlog assume $\mu = 1$. For fixed $\sigma,a$ as above, define $\alpha \stackrel{\rm def}{=}\frac{a+1}{2}\in(a,1)$, and let $\beta$ be the solution of the equation $$
\sigma^2 + 1 = \alpha + \beta - \alpha\beta
$$
i.e. $\beta = 1+\frac{\sigma^2}{1-\alpha} > 1$.
Let $X$ be the random variable taking values in $\{\alpha,\beta\}$, defined as
$$
X = \begin{cases}
\alpha &\text{ w.p. } \frac{\beta-1}{\beta-\alpha} \\
\beta &\text{ w.p. } \frac{1-\alpha}{\beta-\alpha} \\
\end{cases}
$$
so that indeed
$$
\begin{align}
\mathbb{E} X &= 1 \\
\operatorname{Var} X &= - \mathbb{E}[X^2]- (\mathbb{E} X)^2 \\
&= \frac{1}{\beta-\alpha}\left(\alpha^2(\beta-1) + \beta^2(1-\alpha)\right)- 1 = \alpha+\beta-\alpha\beta -1 \\
&= \sigma^2.
\end{align}
$$
$X$ satisfies all the assumptions, and yet
$$
\mathbb{P}\{X \leq a\} = \mathbb{P}\{X < \alpha\} = 0.
$$
At that point, it looks to me that one would need the assumption that $X$ be bounded to get some interesting lower bound.
Best Answer
let $a:=\log\left(\frac{1}{1-x}\right)$. Then
$$(1-x)^r=\frac{1}{e^{ra}}=\frac{1}{1+ra+\frac{a^2r^2}{2!}+\frac{a^3r^3}{3!}+\ldots}$$
Now truncate the infinite series in the denominator to your heart's content. For example
$$(1-x)^r\leq \frac{1}{1+ra}$$.