<s>[INST] Stuck with modular arithmetic problem using multiplication property I have the following problem:

Given $k\geq 1$, find $h$ such that
$$2^h \frac{4^k-1}{3}-1 \equiv 0 ~(\text{mod}~3).$$

This is my attempt using the invariance of multiplication:
$$2^h \frac{4^k-1}{3} \equiv 1 ~(\text{mod}~3) \Rightarrow 2^h \frac{4^k-1}{3}\cdot3 \equiv 1\cdot3 ~(\text{mod}~3) \Rightarrow 2^h(4^k-1) \equiv 0 ~(\text{mod}~3)$$
Since I know that
$$4^k - 1 \equiv 0 ~(\text{mod}~3) $$
then I can conclude that 
$$\forall k, \forall h, 2^h(4^k-1) \equiv 0 ~(\text{mod}~3)$$
Anyway, this seems really wrong, since if I choose $h=3$ and $k=1$, then I get:
$$2^3 \frac{4^1-1}{3}-1 = 7 \not\equiv 0 ~(\text{mod}~3).$$
What's wrong with this proof?
 [/INST] $$
\begin{align}
\frac{(1+3)^k-1}{3}
&=\frac{\left[\binom{k}{0}3^0+\binom{k}{1}3^1+3^2\sum\limits_{j=2}^k\binom{k}{j}3^{j-2}\right]-1}{3}\\
&=\binom{k}{1}+3\sum\limits_{j=2}^k\binom{k}{j}3^{j-2}\\
&\equiv\binom{k}{1}\pmod3
\end{align}
$$
Therefore,
$$
2^h\frac{(1+3)^k-1}{3}-1\equiv0\pmod3
$$
is the same as
$$
2^hk\equiv1\pmod3
$$
which is solvable when $k\not\equiv0\pmod3$ and then
$$
h=\left\{\begin{array}{}
0&\text{if }k\equiv1\pmod3\\
1&\text{if }k\equiv2\pmod3\\
\end{array}\right.
$$
satisfies the equation.
</s>