Is it possible to simplify $n \equiv - \frac d 4\bigg(\frac 3 2\bigg)^{m-2} - 2 \pmod {3^m}$ further? Is it possible to simplify this further:
$$
n \equiv - \frac d 4\bigg(\frac 3 2\bigg)^{m-2} - 2 \pmod {3^m} \qquad \text{where } n, d \geq 0 \text{ and } m \geq 2 \text{ are integers}
$$
or to glean some info on what $n$ should be to satisfy the equivalence?

I obtained the above as follows:
I want to know when
$$
\frac 1 {3^m} \big(2^m(n+2) - (2 - \Delta) 3^m \big)
\qquad
\text{where } \Delta = \frac d 9 \text{ for some integer } d \geq 0
$$
is an integer. We have
$$
\begin{align}
2^m (n+2) &\equiv (2 - \Delta) 3^m &\pmod {3^m}\\
2^m (n+2) &\equiv (2 - \frac d 9) 3^m &\pmod {3^m}\\
2^m (n+2) &\equiv 2 \cdot 3^m - \frac d {3^2} \cdot 3^m &\pmod {3^m}\\
2^m (n+2) &\equiv - d \cdot 3^{m-2} &\pmod {3^m}\\
4 \cdot 2^{m-2} (n+2) &\equiv - d \cdot 3^{m-2} &\pmod {3^m}\\
n+2 &\equiv - \frac d 4 \bigg(\frac 3 2\bigg)^{m-2} &\pmod {3^m}\\
n &\equiv - \frac d 4 \bigg(\frac 3 2\bigg)^{m-2} - 2 &\pmod {3^m}
\end{align}
$$
From $2^m (n+2) \equiv - d \cdot 3^{m-2} \pmod {3^m}$, I know that if $d \equiv 0 \pmod 9$, then $n \equiv -2 \pmod {3^m}$.
Also, by writing $d = 9k + (d \bmod 9)$, we have
$$
\begin{align}
2^m (n+2) &\equiv - (9k + (d \bmod 9)) \cdot 3^{m-2} &\pmod {3^m}\\
2^m (n+2) &\equiv -9k \cdot 3^{m-2} + -(d \bmod 9) \cdot 3^{m-2} &\pmod {3^m}\\
2^m (n+2) &\equiv -k \cdot 3^m + -(d \bmod 9) \cdot 3^{m-2} &\pmod {3^m}\\
&\boxed{
2^m (n+2) \equiv -(d \bmod 9) \cdot 3^{m-2} \pmod {3^m}
}
\end{align}
$$

Simplfy $n \equiv - \dfrac d 4\bigg(\dfrac 3 2\bigg)^{m-2} - 2 \pmod {3^m} \qquad \text{where } n, d \geq 0 \text{ and } m \geq 2 \text{ are integers}$.
Multiplying by $\dfrac94$ we get
$$\dfrac94n\equiv -\dfrac d4 \left(\dfrac 32\right)^m-\dfrac 92\pmod{3^m}\Rightarrow 9n\equiv \left(\dfrac{-d}{2^m}\right)3^m-18\pmod{3^m}$$ Then $$\boxed{n\equiv-2\mod{3^m}}$$
