How can we prove that there are no other integers with $\phi(n)=2$ besides 3,4,6?

The trickiest thing about this proof is figuring out how to organize the casework. Here's one way to do it:

If $n = 2^k$ is a power of two then $\varphi(n) = 2^{k-1}$ so we see that we can only have $k = 2$, so $n = \boxed{4}$.

Otherwise, $n$ has some odd prime power factor $p^k$, and then $\varphi(n)$ must be divisible by $\varphi(p^k) = (p-1) p^{k-1}$. Since $p$ is odd, $2 \mid (p-1)$, so $\varphi(p^k)$ will be strictly larger than $2$ unless $p = 3, k = 1$. So now we must have $n = 2^k \cdot 3$, which gives $\varphi(n) = 2^k$ for $k \ge 1$, hence $k = 1$, so $n = \boxed{6}$, or $\varphi(3) = 2$ for $k = 0$, so $n = \boxed{3}$.

Exercise. Generalize this argument to show that for any $m$ there are finitely many $n$ such that $\varphi(n) = m$. Can you compute which $n$ these are for other small values of $m$, say $m = 4$ or $m = 6$? (Hint: prove it in two stages. First prove that there are only finitely many possible prime factors for $n$. Second prove that the exponent of each of these prime factors is bounded. Working through the small cases $m = 4$ and $m = 6$ first would be a good idea as a warmup.)