$coNP$ and $\oplus P$?
The result that $\mathrm{MOD}_{2^m}\mathrm P=\mathrm{MOD}_2\mathrm P$ only applies when $m$ is constant. Thus, your strategy only works if $t$ is constant, which is a trivial case. You would need it to apply to $t\approx n^c$. (Actually, the argument works for $m=O(\log n)$, but this is still not enough.)
Specifically, the reduction of $\mathrm{MOD}_{2^m}\mathrm P$ to $\mathrm{MOD}_2\mathrm P$ relies on the simple number-theoretic fact that
$$2^m\mid s\iff2\mid s\land2\mid\binom s2\land2\mid\binom s{2^2}\land\dots\land2\mid\binom s{2^{m-1}}.$$
If $s$ counts the number of objects satisfying some property, then $\binom s{2^r}$ counts the number of $2^r$-element sets of objects with that property.
That is, given a language $L\in\mathrm{MOD}_{2^m}\mathrm P$ of the form
$$w\in L\iff2^m\mid\#\{x&lt;2^{n^c}:R(w,x)\},$$
where $n$ denotes $|w|$ and $R\in\mathrm P$, we can express $L$ as
$$w\in L\iff\forall r&lt;m\ 2\mid\#\{(x_0,\dots,x_{2^r-1})\in[0,2^{n^c})^{2^r}:R_r(w,x_0,\dots,x_{2^r-1})\},$$
where
$$R_r(w,x_0,\dots,x_{2^r-1})\iff x_0&lt;\dots&lt;x_{2^r-1}\land\forall i&lt;2^r\:R(w,x_i).$$
(These $m$ instances of $\oplus\mathrm P$ can be then combined to a single $\oplus\mathrm P$ predicate as $\oplus\mathrm P$ is closed under poly-time Turing reductions; in fact, even $\mathrm{\oplus P^{\oplus P}=\oplus P}$.)
As long as $m=O(\log n)$, the sequences $(x_0,\dots,x_{2^r-1})$ can be represented with $n^{O(1)}$ bits, and the predicates $R_r$ are computable polynomial time, but neither is true for larger $m$.