How to prove that $\int _0^\infty\frac{\text{arcsinh}^nx}{x^m}dx$ is a rational combination of zeta values?

Let me first establish the case $m=2$. The general case is established in the "Added" section below.

By making the substitution $x=\sinh t$ we get \begin{align*} I(n,2)&=\int_0^\infty t^n\frac{\cosh t}{\sinh^2 t}\,dt =2\int_0^\infty t^n\frac{e^t+e^{-t}}{(e^t-e^{-t})^2}dt\\ &=2\int_0^\infty t^n\left(\sum_{r=0}^\infty(2r+1)e^{-(2r+1)t}\right)dt\\ &=2\sum_{r=0}^\infty(2r+1)\left(\int_0^\infty t^n e^{-(2r+1)t}dt\right)\\ &=2\sum_{r=0}(2r+1)^{-n}\left(\int_0^\infty t^n e^{-t}dt\right)\\ &=2n!\sum_{r=0}(2r+1)^{-n}=2n!(1-2^{-n})\zeta(n)=\frac{n!}{2^{n-1}} \bar\zeta(n). \end{align*}

Added. Here is the general case for $m=2k$ even. In the umbral notation, we need to replace $(\bar Z/2)^r$ by $$ 2^{-r}\bar\zeta (r)=(1-2^{-r})\zeta(r)=\sum_{s\text{ odd}}s^{-r}, $$ hence the conjectured formula really means $$ I(n,2k)\overset{?}=\frac{2n!}{(2k-1)!}\sum_{s\text{ odd}}s^{-n}\prod_{j=-k+1}^{k-2}\bigl(s-(2j+1)\bigr).$$ The inside product is a polynomial of degree $2k-2$, and we see that it vanishes for $s\leq 2k-3$. Hence we make the substitution $s=2r+2k-1$, and we rewrite the conjectured formula as $$ I(n,2k)\overset{?}=n!\sum_{r=0}^\infty (2r+2k-1)^{-n} \frac{2^{2k-1}}{2k-1}\binom{r+2k-2}{2k-2}.$$ We prove this equation. From the definition, similarly as in the basic case $k=1$ above, we find \begin{align*} I(n,2k)&=2^{2k-1}\int_0^\infty t^n e^{(1-2k)t}(1+e^{-2t})(1-e^{-2t})^{-2k}dt\\ &=2^{2k-1}\int_0^\infty t^n e^{(1-2k)t} \sum_{r=0}^\infty\binom{-2k}{r}(-1)^r \left(e^{-2rt}+e^{-2(r+1)t}\right)dt\\ &=2^{2k-1}\int_0^\infty t^n e^{(1-2k)t}\left(1+\sum_{r=1}^\infty\left(\binom{-2k}{r}-\binom{-2k}{r-1}\right)(-1)^r e^{-2rt}\right)dt\\ &=2^{2k-1}\int_0^\infty t^n e^{(1-2k)t}\left(1+\sum_{r=1}^\infty\left(\binom{r+2k-1}{r}+\binom{r+2k-2}{r-1}\right)e^{-2rt}\right)dt\\ &=2^{2k-1}\int_0^\infty t^n \sum_{r=0}^\infty\left(\binom{r+2k-1}{2k-1}+\binom{r+2k-2}{2k-1}\right)e^{-(2r+2k-1)t}dt\\ &=2^{2k-1}\sum_{r=0}^\infty \left(\binom{r+2k-1}{2k-1}+\binom{r+2k-2}{2k-1}\right) \int_0^\infty t^ne^{-(2r+2k-1)t}dt\\ &=2^{2k-1}\sum_{r=0}^\infty \left(\binom{r+2k-1}{2k-1}+\binom{r+2k-2}{2k-1}\right)\frac{n!}{(2r+2k-1)^{n+1}}\\ &=2^{2k-1}\sum_{r=0}^\infty \frac{2r+2k-1}{2k-1}\binom{r+2k-2}{2k-2}\frac{n!}{(2r+2k-1)^{n+1}}\\ &=n!\sum_{r=0}^\infty (2r+2k-1)^{-n}\frac{2^{2k-1}}{2k-1}\binom{r+2k-2}{2k-2}. \end{align*} The proof is complete for the case of even $m$.

The case of odd $m$ is similar, so I omit the details.

Letting $y=\text{arcsinh}\, x$ and integrating by parts, we have $$I(n,m)=\frac n{m-1}\,J(n-1,m-1) $$ if $n\ge m\ge2$, where $$J(p,q):=\int_0^\infty\frac{y^p}{\sinh^qy}\,dy.$$ Formula 1.4.24.1 in Prudnikov--Brychkov--Marichev (PBM, Vol. 1, ISBN 5-9221-0323-7) implies
$$J(p,q)= \frac{p(p-1)}{(q-1)(q-2)}\,J(p-2,q-2) -\frac{q-2}{q-1}\,J(p,q-2) $$ if $p\ge q>2$.
That formula in PBM (which is reproduced in my answer at Is there a closed form of $\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$?) should be easy to obtain/check. Thus, we have a recursion to reduce the value of $I(n,m)$ to those of $J(p,1)$ and $J(p,2)$, that is, to those of $I(n,2)$ and $I(n,3)$.