I have put in a full proof for completeness. Claim 2 below says that the order of $3$ in $(\mathbb{Z}/2^{\ell+2}\mathbb{Z})^*$ is
exactly $2^{\ell}$ as you have noted. It sounds that you have already observed Claim 2 yourself but I included this for anyone else who is interested. Then Claim 3 stated and established below is the crux on how you finish establishing that the stated set is indeed a reduced residue system. The Lifting The Exponent Lemma is not needed!
THM 1: Let $\ell$ be a positive integer. Then the set $\{\pm 3^k;
k \in \{1,2,\ldots, 2^{\ell}\}\}$ form a reduced residue mod $2^{\ell+2}$.
We prove THM 1 via Claims 2 and 3 stated below:
Claim 2: Let $\ell$ be a positive integer. Let $r_{\ell}$ be
the smallest positive integer such that the equation
$3^{r_{\ell}}\equiv_{2^{\ell+2}} 1$ holds. Then the equation
$r_{\ell} = 2^{\ell}$ holds. Equivalently, for all positives integers $k$ in $\{1,2, \ldots, 2^{\ell}-1\}$, the relation
$3^{k} \not \equiv_{2^{\ell+2}} 1$ holds.
Proof of Claim 2: We first note the following: First, $r_{\ell}$
must be a power of $2$ [as $|(\mathbb{Z}/2^{\ell'}\mathbb{Z})^*|$ is
a power of $2$, and $3$ is in
$(\mathbb{Z}/2^{\ell'}\mathbb{Z})^*$. Next, $r_{\ell+1} \ge r_{\ell}$
for each positive integer $\ell$. [Indeed, if for any positive
integers $\ell'$ and $r$ the equation $3^r \equiv_{2^{\ell'+1}} 1$
holds, then so does the equation $3^r \equiv_{2^{\ell'}} 1$.] So to
finish the proof of Claim 2, it suffices to do two things:
(A) First, for each positive integer
$\ell$, show that the equation $3^{2^{\ell+1}} \equiv_{2^{\ell+3}} 1$ is
satisfied.
(B) Next, show that the relation
$3^{2^{r+1}} \not \equiv_{2^{\ell+4}} 1$ is also satisfied.
However, both (A) and (B) are clearly true for $\ell=1$. Thus, use
induction. In particular, assume that the equation
$3^{2^{\ell}} \equiv_{2^{\ell+2}} 1$ holds, as well as the relation
$3^{2^{\ell}} \not \equiv_{2^{\ell+3}} 1$, or equivalently,
$3^{2^{\ell}} \equiv_{2^{\ell+2}} 1 + C2^{\ell+2}$, for some odd
integer $C$. Then these imply:
$$3^{2^{\ell+1}} = (3^{2^{\ell}})^2$$ $$= (1+C2^{\ell+2})^2$$ $$= 1 +[2C + C^22^{\ell+2}]2^{\ell+2}$$
$$ = 1+[C+C^22^{\ell+1}]2^{\ell+3},$$
which gives indeed:
The equation $3^{2^{r+1}} \equiv_{2^{\ell+3}} 1$ is
satisfied.
The relation $3^{2^{r+1}} \not \equiv_{2^{\ell+4}} 1$ is also
satisfied. [Indeed, $C+C^22^{\ell+1}$ as above is an odd integer.]
Thus, we have done (A) and (B) as above, and as already noted, this
suffices to establish Claim 2. $\surd$
Claim 3: Let $\ell$ be a positive integer. There is no integer
$k \in \{0,1,2, \ldots, 2^{\ell}-1\}$ such that $3^k
\equiv_{2^{\ell+2}} \ -1$ holds.
Proof of Claim 3: If $\ell \le 3$ then this can be checked
directly. Then if there is a
$k \in \{0,1, \ldots, 2^{\ell}-1\}$ such that $3^k \equiv_{2^{\ell+2}} \ -1$,
then $k$ must be positive, and then both the relation $2k < 2^{\ell+1}$, and the equation $3^{2k} \equiv_{2^{\ell+3}} 1$ hold. This contradicts Claim 2 above. And thus Claim 3 follows. $\surd$
We now finish the proof of THM 1. To establish THM 1, it suffices to show both that $3^{k_1} \not \equiv_{2^{\ell+2}} 3^{k_2}$ for distinct integers $k_1,k_2 \in \{1,2, \ldots, 2^{\ell}\}$, and
$3^{k_1} \not \equiv_{2^{\ell+2}} -3^{k_2}$ for integers $k_1,k_2 \in \{1,2, \ldots, 2^{\ell}\}$. We do this next.
Let us suppose that there are
distinct integers $k_1,k_2 \in \{1,2, \ldots, 2^{\ell}\}$ such that
$3^{k_1} \equiv_{2^{\ell+2}} 3^{k_2}$. Then assuming WLOG $k_1 >
k_2$, this gives the equation $$3^{k_1-k_2} \equiv_{2^{\ell+2}} 1.$$
[make sure you see why.] This is impossible via Claim 2 [because $k_1-k_2$ is an integer in $\{1,2,\ldots, 2^{\ell}-1\}$].
Now suppose
that there are integers $k_1,k_2 \in \{1,2, \ldots, 2^{\ell}\}$ such that $3^{k_1} \equiv_{2^{\ell+2}} -3^{k_2}$, with $k_1 \ge k_2$. Then this gives the equation $$3^{k_1-k_2} \equiv_{2^{\ell+2}} \ -1.$$ This is
impossible via Claim 3 [because $k_1-k_2$ is an integer in $\{0,1,2,\ldots, 2^{\ell}-1\}$].
Thus as already noted, from this THM 1 follows.
