Unfortunately, I don't see any direct way to use the approach you stated in your update. Instead, here is a way to use the provided hint. Have there be $k$ (starting with $k = 0$) distinct, odd primes $p \in T$, with these primes being $p_i$ for $1 \le i \le k$. Then define
$$n = \prod_{i=1}^{k}\varphi(p_i - 1) \tag{1}\label{eq1A}$$
By Euler's theorem, this gives
$$3^{n} \equiv 1 \pmod{p_i - 1} \; \forall \; 1 \le i \le k \tag{2}\label{eq2A}$$
Thus, we have
$$u = 2^{3^{n}} - 3 \equiv 2 \cdot 2^{3^n-1}-3 \equiv 2 - 3 \equiv -1 \pmod{p_i} \; \forall \; 1 \le i \le k \tag{3}\label{eq3A}$$
Next, due to $u \equiv 2 \pmod{3}$, there's at least one prime which is $\equiv 2 \pmod{3}$ which divides $u$. Choose any one of these (e.g., the smallest one) and call it $q$. Since \eqref{eq3A} shows none of the $p_i$ divide $u$, then $q$ must be different from all of them.
Since $q \equiv 2 \pmod{3}$, then $\gcd(3, q - 1) = 1$. Thus, there exist multiplicative inverses for $3^{n}$ modulo $q - 1$, so pick the smallest positive one and call it $m$. This therefore gives, for some integer $s$,
$$3^{n}(m) \equiv 1 \pmod{q - 1} \; \; \to \; \; 3^{n}(m) = (q - 1)s + 1 \tag{4}\label{eq4A}$$
Next, using this gives that
$$\begin{equation}\begin{aligned}
2^{3^{n}} & \equiv 3 \pmod{q} \\
\left(2^{3^{n}}\right)^{m} & \equiv 3^{m} \pmod{q} \\
2^{(q-1)s + 1} & \equiv 3^{m} \pmod{q} \\
\left(2^{q-1}\right)^{s}(2) & \equiv 3^{m} \pmod{q} \\
2 & \equiv 3^{m} \pmod{q}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
As you stated in the question, this is sufficient. Nonetheless, to prove that $q \in T$, have the set of remainders mod $q$ of $\{2, 2^2, 2^3, \ldots \}$ be $U$ and of $\{3, 3^2, 3^3, \ldots \}$ be $V$. It is relatively easily to show $U \subseteq V$ and $V \subseteq U$, so $U = V$.
Thus, set $p_{k+1} = q$, increment $k$ and repeat the above steps. Doing this infinitely often shows there are an infinite number of distinct primes $p_{i} \in T$.
Best Answer
First, note that for any $r$ there at most two numbers $s$ satisfying
This is because the only possible options are $r+1$ or $\frac{pr}{p-1}$ where $p$ is the largest prime dividing $r$.
Second, we can write $n=s_1s_2\cdots s_k$ where each $s_i$ is a power of a different prime. Then if $\phi(s_i)=r_i$ for each $i$, we have $\phi(n)=r_1r_2\cdots r_k$ and we have at most one $i$ with $r_i=1$ (since this only occurs if $s_i=2$). There are only finitely many ways to write $m$ as a product of some positive integers $r_i$ with at most one $r_i=1$, and for each such product there are finitely many ways to reconstruct the $s_i$.
For the second question, numbers which don't appear as the totient function of other numbers are called "nontotients" - see the wikipedia article on them for more details.