If $\varphi(x) = m$ has exactly two solutions is it possible that both solutions are even?
Here, $\varphi(x)$ is Euler's phi function, the number of positive integers less than or equal to $x$ that are relatively prime to $x$.
It appears that when $\varphi(x) = m$ has exactly two solutions then one of the solutions, $x$, is odd and the other solution is even, specifically $2\times x$. The first few integers $x$ such that $\varphi(x) = m$ has exactly two solutions are: $1,11,23,29,31,47,53,81,59,\dots$ where the other solution is necessarily $2\times x$. For example, $\varphi(81) = \varphi(162) = 54$ and there are no other integers $k$ such that $\varphi(k) = 54$. See the sequence $A007366$ in Sloane's OEIS. I determined that the initial terms of this sequence were all odd with a brute force Mathematica code that is accurate (and timely) for about $1000$ terms.
The emperical evidence suggests that if $\varphi(x) = m$ has exactly two solutions then exactly one of them is odd. I want to prove this statement.
It is straight forward to show that both solutions cannot be odd since if $x$ is odd then $\varphi(x) = \varphi(2\times x)$.
I am considering the case where both solutions are even, hoping for a contradiction. I have determined that if both solutions, say $x$ and $y$, are even then $x$ and $y$ are both divisible by $4$. Also, in this case, at least one of $x$ or $y$ is divisible by $3$. Now I am stuck. Is there some way to reach a contradiction here.
Is there another way to prove (or attempt to prove) the conjecture?