For which integer values of $m \geq 0$ and $n \geq 0$ is $A_m(n) = (\frac 2 3) ^ m (n + 2) - 2$ a positive integer? For which integer values of $m \geq 0$ and $n \geq 0$ is $A_m(n) = (\frac 2 3) ^ m (n + 2) - 2$ a positive integer?
I made a table of the first few expressions for $m \in [0, 5]$
$$
\begin{align}
m && A_m(n) = (\frac 2 3) ^ m (n+2) - 2 && \text{integer iff}\\
0 && n && \text{always}\\
1 && \frac 2 3 (n - 1) && n \equiv 1 \pmod 3\\
2 && \frac 1 9 (4n - 10) && 4n \equiv 10 \equiv 1 \pmod 9\\
3 && \frac 1 {27} (8n - 38) && 8n \equiv 38 \equiv 11 \pmod {27}\\
4 && \frac 1 {81} (16n - 130) && 16n \equiv 130 \equiv 49 \pmod {81}\\
5 && \frac 1 {243} (32n - 422) && 32n \equiv 422 \equiv 179 \pmod {243}
\end{align}
$$
In general,
$$
A_m(n) = (\frac 2 3) ^ m (n + 2) - 2 = \frac 1 {3^m} (2^m n - 2(3^m - 2^m))
$$
so $A_m(n)$ is an integer when
$$
\begin{align}
2^m n &\equiv 2(3^m - 2^m) &\pmod {3^m}\\
2^{m-1} n &\equiv 3^m - 2^m &\pmod {3^m}\\
2^{m-1} n &\equiv -(2^m) &\pmod {3^m}\\
2^{m-1} n &\equiv -2(2^{m-1}) &\pmod {3^m}\\
n &\equiv -2 &\pmod {3^m}\\
n &\equiv 3^m - 2 &\pmod {3^m}
\end{align}
$$
Is this correct? I'm not $100%$ on the rules of manipulating modular equivalencies. Thanks for any help!

Background
Let $A(n) = \frac 2 3 (n - 1) = \frac 2 3 n - \frac 2 3$.
Let $A_m(n) = \underbrace{A(A(A(...A}_{m \text{ times}}(n)..)))$. For example, $A_3(n) = A(A(A(n)))$. Then,
$$
\begin{align}
A_0(n) &= n\\
A_1(n) &= \frac 2 3 n - \frac 2 3\\
A_2(n) &= \frac 2 3 (\frac 2 3 n - \frac 2 3) - \frac 2 3 = (\frac 2 3)^2 n - (\frac 2 3)^2 - \frac 2 3\\
A_3(n) &= (\frac 2 3)^3 n - (\frac 2 3)^3 - (\frac 2 3)^2 - \frac 2 3
\end{align}
$$
In general,
$$
\begin{align}
A_m(n) &= (\frac 2 3)^m n - \sum_{k=1}^m (\frac 2 3)^k\\
&= r^m n - \frac {r - r^{m+1}} {1 - r} &\text{Geometric series with } r = \frac 2 3\\
&= r^m n - r\frac {1 - r^m} {\frac 1 3}\\
&= r^m n - 3r(1 - r^m)\\
&= r^m n - 2(1 - r^m)\\
&= r^m n - 2 + 2r^m\\
&= r^m (n + 2) - 2\\
&= (\frac 2 3)^m (n + 2) - 2\\
&= \frac 1 {3^m} (2^m n - 2(3^m - 2^m))
\end{align}
$$

$\left(\dfrac 2 3\right) ^ m (n + 2) - 2=k\iff 2^m(n+2)=3^m(k+2)$ so if you take
$$n+2=3^m\\k+2=2^m$$ and you get infinitely many solutions..
