This was intended as an extended comment but started to have too many formulas, so I thought that it would be more legible if posted as an answer.
(You don't actually state if there are $\nu$-null sets which are not Lebesgue null, so I'm going to build an example which is mutually absolutely continuous wrt Lebesgue measure.)

If $h$ is unbounded then it seems to me that your shift operator could be unbounded. You don't specify whether you want $p$ to be in the reflexive range, so let me take $p=2$ just to be sure, and take $n=1$, $a=1$ for simplicity.
Take $h(x)=|x|^{-3/4}$ for $x\in [-1,1]$ and $h(x)=e^{-|x|}$ outside that interval, so that $h \in L^1_m({\bf R})$. Put $d\nu = h\ dm$, so that $\nu$ is a finite measure on ${\bf R}$.

Consider
$$ f(x) = \begin{cases}(x-1)^{-1/3} & \hbox{if $x\in (1,2]$} \\ 0 & \hbox{otherwise} \end{cases}$$
This belongs to $L^2_\nu({\bf R})$ since
$$ \int_{\bf R} |f(x)|^2 h(x) \,dx = \int_1^2 (x-1)^{-2/3} e^{-x}\,dx <\infty $$
On the other hand,
$$ (T_1f)(x) =f(x+1) = \begin{cases} x^{-1/3} & \hbox{if $x\in (0,1]$} \\ 0 & \hbox{otherwise} \end{cases}$$
so
$$ \int_{\bf R} |T_1f(x)|^2 h(x) \,dx = \int_0^1 x^{-2/3} x^{-3/4}\,dx = +\infty $$