Suppose that $\varphi(n) = 2^m$ for some $m \in \mathbb{Z}^+$. Based on the formula of $\varphi(n)$, we have
$$\varphi(n) = \prod_{p|n} p^{t-1}(p-1) = 2^m$$
with $v_p(n) = t$. This implies that $t = 1$ for all odd prime divisors, since otherwise, $p \mid 2^m$, a contradiction. Thus $p-1$ is a power of 2.

With odd prime divisor of $n$, put $p = 2^s + 1$ for some $s \in \mathbb{Z}^+$. If $s = 1$ then $p = 3$, otherwise $s$ must be even since $p \equiv (-1)^s + 1 \equiv 0 \pmod{3}$ when $s$ is odd, contradiction. In case $s$ has some proper odd prime divisor $q$ then put $s = qt$ with $t > 1$ then $2^s + 1 = (2^t)^q + 1$ is divisible by $2^t + 1$, contradiction also.

These imply that $s$ is also a power of 2, so by putting $s = 2^k$, we get $p = 2^{2k} + 1$. Now let $n = 2^a p_1 p_2 \dots p_l$ with $a \ge 0, l \ge 0$ and $p_1, p_2, \dots, p_l$ are odd primes in ascending order, in which $p_1 = 2^{2k} + 1$. So
$$2n = \sigma(n) = (2^{a+1} - 1)(p_1 + 1)(p_2 + 1)\dots(p_l + 1)$$
is the sum of divisors of $n$. Note that $l > 0$, otherwise, $\sigma(n) = 2^{a+1}-1$, contradiction. If $p_1 > 3$ then $3 \nmid 2n$ and
$$p_1 + 1 = 2^{2k} + 2 = 2(2^{2k-1} + 1) \equiv 2((-1)^{2k-1} + 1) \equiv 0 \pmod{3}$$
implying that $3 \mid 2n$, contradiction. Thus $p_1 = 3$ and $v_3(2n) = 1$ leads to $l \le 2$. There are two cases

* If $l = 2$ then $2^{a+1} \cdot 3 \cdot p_2 = (2^{a+1} - 1) \cdot 4 \cdot (p_2 + 1)$, note that $v_2(p_2+1) = 1$ so comparing the exponent of 2 in both sides to get $a = 2$. Thus $6p_2 = 7(p_2+1)$, contradiction.
* If $l = 1$ then $2^{a+1} \cdot 3 = (2^{a+1} - 1) \cdot 4$, then $a = 1$ implies that $n = 6$ is a perfect square.

Therefore, $n = 6$ is only solution of this problem.