Consider the integral of the general form
$$
J_{n,m} = \int_0^\infty x^n \text{csch}^{m}x \>\mathrm{d}x
$$
and note that
$$
I_n = \int_0^\infty \Big(\frac{x}{\sinh x}\Big)^n d x= J_{n,n}
$$
As shown below, $ J_{n,n}$ can be computed via the reduction formula
$$J_{n,m}=\frac{n(n-1)}{(m-1)(m-2)}J_{n-2,m-2}
- \frac{m-2}{m-1} J_{n,m-2}\tag1
$$
along with the starting values
$$J_{2k-1,1}= \frac{(2^{2k}-1)(2k-1)!}{2^{2k-1}}\zeta(2k),\>\>\>\>\>
J_{2k,2}=\frac{k(2k-1)!}{2^{2k-2}}\zeta(2k)\tag2
$$
$$\zeta(2)=\frac{\pi^2}6, \>\>\>\zeta(4)= \frac{\pi^4}{90},\>\>\>
\zeta(6)=\frac{\pi^6}{945},\>\>\>\zeta(8)= \frac{\pi^8}{9450}, \>\>\>\cdots
$$
Proof: Apply
$$\int \text{csch}^{m}x\>\mathrm{d}x
= -\frac{\text{csch}^{m-2}x \>{\coth}x }{m-1} -\frac{m-2}{m-1} \int \text{csch}^{m-2}x\>\mathrm{d}x
$$
to integrate $J_{n,m}$ by parts
$$J_{n,m} = \frac{n}{m-1} \int x^{n-1} \text{csch}^{m-2}x \coth x \mathrm{d}x- \frac{m-2}{m-1}J_{n,m-2}\tag3
$$
where
$$ \int x^{n-1} \text{csch}^{m-2}x \coth x \mathrm{d}x
=-\int x^{n-1} \text{csch}^{m-3}x \>d(\text{csch}x)\overset{IBP} = \frac{n-1}{m-2}
I_{n,m-2} $$
Plug into (3) to obtain the reduction formula (1). To evaluate
$J_{2k-1,1}$ and $ J_{2k,2}$ given in (2), apply the substitution $x=-\ln t$. Then
$$J_{2k,2}= 4\int_0^1 \frac{t \ln^{2k}t}{(1-t^2)^2}dt\overset{t^2\to t}=\frac1{2^{2k-1} }\int_0^1 \frac{\ln^{2k}t}{(1-t)^2}dt\\
\overset{\text{IBP}}=-\frac k{2^{2k-2} }\int_0^1 \frac{\ln^{2k-1}t}{1-t}dt
=\frac{k(2k-1)!}{2^{2k-2}}\zeta(2k)
$$
$$J_{2k-1,1}= -2\int_0^1 \frac{\ln^{2k-1}t}{1-t^2}dt
=\frac{(2^{2k}-1)(2k-1)!}{2^{2k-1}}\zeta(2k)
$$
Displayed below are some results via the reduction formulae (1) and (2):
\begin{align}
I_1 =J_{1,1} &= \frac32 \zeta(2)= \frac{\pi^2}4\\
I_2 =J_{2,2} &= \zeta(2)= \frac{\pi^2}6\\
I_3 =J_{3,3} &= 3J_{1,1}-\frac12 J_{3,1} = 3\cdot \frac{\pi^2}4 - \frac12 \cdot \left(\frac{45}4 \zeta(4)\right)
= \frac{3\pi^2}4 - \frac{\pi^4}{16}\\
I_4 =J_{4,4} & = 2J_{2,2}-\frac23 J_{4,2} = 2\cdot \frac{\pi^2}6 - \frac23 \cdot \left(3\zeta(4)\right)
= \frac{\pi^2}3 - \frac{\pi^4}{45}\\
I_5 =J_{5,5}
& = \frac53 J_{3,3}-\frac34\left( 10J_{3,1} -\frac12J_{5,1} \right)
= \frac{5\pi^2}4 - \frac{25\pi^4}{24} +\frac{3\pi^6}{32}\\
I_6 =J_{6,6}
&= \frac32 J_{4,4}-\frac45 \left( 5J_{4,2} -\frac23 J_{6,2} \right)
= \frac{\pi^2}2 - \frac{\pi^4}{6} +\frac{4\pi^6}{315} \\
I_7 =J_{7,7} &=\>\cdots
\end{align}