You cannot pick an alternative to $n=2^j$ out of thin air and work only on that alternative. What if $n=7$? What if $n=573428901367503218567329178542$? You never considered those.
Write $n=p_1^{a_1}\cdots p_r^{a_r}$, as the prime factorization of $n$, with $p_1\lt p_2\lt\cdots\lt p_r$ distinct primes, $a_i\gt 0$ for each $i$. Then $\phi(n) = (p_1-1)p_1^{a_1}\cdots (p_r-1)p_r^{a_r-1}.$ If $r\gt 1$, then the largest power of $p_r$ that divides $2\phi(n)$ is $a_r-1$ (prove it; it takes half a line), while the largest power of $p_r$ that divides $n$ is of course $a_r$. Since $a_r-1\neq a_r$, we get a contradiction. Thus, $r=1$, so $n$ is a prime power.
Therefore, $n=p_1^{a_1}$, $\phi(n) = (p_1-1)p_1^{a_1-1}$, and $p_1^{a_1} = 2(p_1-1)p_1^{a_1-1}$. Therefore...