I've been trying to prove this inequality for a while.
$$\int_{0}^{\infty}{e^{-2(m-2)s}(1+s)^m ds}\leq\frac{1}{m-3}\text{ for }m>3.$$ It can be rewritten in terms of the incomplete gamma function $$\Gamma(m+1,2(m-2))\leq\frac{(2(m-2))^{m+1}e^{-2(m-2)}}{m-3}$$ or the generalized exponential integral $$E_{-m}(2(m-2))\leq\frac{e^{-2(m-2)}}{m-3}.$$
For any version, integrating by parts gives a recurring expression, but it quickly becomes intractable.
The inequality can be directly checked for low values of $m$. It can also be proved that $(m-3)\int_{0}^{\infty}{e^{-2(m-2)s}(1+s)^m ds}$ converges to one as $m$ grows to infinity. But I have not been able to prove that the latter expression is always increasing in $m$.
Using the fact that $1+s\leq e^s$ holds for every $s\geq 0$, we can see that the left hand side is smaller than $\frac{1}{m-4}$, but I am interested in the tighter bound.
On $(3,\infty)$ we have \begin{align} 0 \leq &(m-3)\int_0^\infty e^{-2(m-2)s} (1+s)^m \, {\rm d}s \\ \stackrel{s=e^{t}-1}{=} &(m-3)\int_{0}^\infty e^{-2(m-2)\left(e^{t}-1\right)+(m+1)t} \, {\rm d}t \\ \leq &(m-3) \int_{0}^\infty e^{-2(m-2)\left(t+\frac{t^2}{2}+\frac{t^3}{6}\right)+(m+1)t} \, {\rm d}t \\ \stackrel{u=(t+1)^3}{=} &(m-3) \, \underbrace{e^{-3}\int_1^\infty \frac{e^{-\frac{m-2}{3}(u-1)+3u^{{1}/{3}}}}{3u^{{2}/{3}}} \, {\rm d}u}_{\equiv I_m} \, . \end{align} Now \begin{align} \lim_{m\rightarrow \infty} (m-3)I_m &= \lim_{m\rightarrow \infty} \frac{m-3}{m-2} \, e^{-3} \sum_{k=0}^\infty \frac{3^k}{k!} \left(\frac{m-2}{3}\right)^{-\frac{k+1}{3}+1} e^{\frac{m-2}{3}} \, \Gamma\left(\frac{k+1}{3} , \frac{m-2}{3}\right) \\ &=1 \end{align} as can be seen by expanding $e^{3u^{1/3}}$.
It therefore remains to show that $(m-3)I_m$ is an increasing function. For that the idea is to remove the $m$-dependent pre-factor before taking the derivative to keep it tractable. In fact we have the identity $$ (m-2)I_m = 1 + e^{-3} \int_1^\infty \left\{ u^{-4/3} - \frac{2 u^{-5/3}}{3} \right\} \, e^{-\frac{m-2}{3}(u-1)+3u^{{1}/{3}}} \, {\rm d}u $$ by partial integration. Hence \begin{align} \frac{\rm d}{{\rm d}m} (m-3)I_m &= \frac{\rm d}{{\rm d}m} (m-2)I_m - \frac{\rm d}{{\rm d}m} I_m \\ &=e^{-3} \int_1^\infty \underbrace{\left\{ \frac{u^{1/3}}{9} - \frac{u^{-1/3}}{3} + \frac{u^{-2/3}}{9} + \frac{u^{-4/3}}{3} - \frac{2 u^{-5/3}}{9} \right\}}_{\equiv f(u)} \, e^{-\frac{m-2}{3}(u-1)+3u^{{1}/{3}}} \, {\rm d}u \\ &\geq 0 \end{align} because $f(u)\geq 0$ which can be seen as follows:
First $f(1)=0$ and $$f'(u) = \frac{u^{-2/3}}{27} + \frac{u^{-4/3}}{9} - \frac{2u^{-5/3}}{27} - \frac{4u^{-7/3}}{9} + \frac{10u^{-8/3}}{27} \geq 0 \, .$$
The latter because $f'(1)=0$ and $f'(\infty)=0$ while $$ f''(u)=-\frac{2u^{-5/3}}{81} - \frac{4u^{-7/3}}{27} + \frac{10u^{-8/3}}{81} + \frac{28u^{-10/3}}{27} - \frac{80u^{-11/3}}{81} = 0 $$ has only $1$ real-valued solution $u_0=z^3>1$ where $z$ is the only real solution of $$ z^5 + z^4 + 7z^3 + 2z^2 + 2z - 40 = 0 $$ which corresponds to a maximum of $f'(u)$, since \begin{align} f'''(u_0)&=\frac{10u_0^{-8/3}}{243} + \frac{28u_0^{-10/3}}{81} - \frac{80u_0^{-11/3}}{243} - \frac{280u_0^{-13/3}}{81} + \frac{880u_0^{-14/3}}{243} \\ &\approx -0.002446492623 < 0 \end{align} with $u_0 \approx 3.101517308$.