If $\varphi(x) = m$ has exactly two solutions is it possible that both solutions are even?

As I requested, here are the first twenty examples of numbers $m$ with two answers to $\phi(x)=m.$ I did not print out when $m+1$ was prime, those are the majority of the $m'$s

54 = 2 * 3^3     81 = 3^4     162 = 2 * 3^4     
110 = 2 * 5 * 11     121 = 11^2     242 = 2 * 11^2     
294 = 2 * 3 * 7^2     343 = 7^3     686 = 2 * 7^3     
342 = 2 * 3^2 * 19     361 = 19^2     722 = 2 * 19^2     
506 = 2 * 11 * 23     529 = 23^2     1058 = 2 * 23^2     
580 = 2^2 * 5 * 29     649 = 11 * 59     1298 = 2 * 11 * 59     
812 = 2^2 * 7 * 29     841 = 29^2     1682 = 2 * 29^2     
930 = 2 * 3 * 5 * 31     961 = 31^2     1922 = 2 * 31^2     
1144 = 2^3 * 11 * 13     1219 = 23 * 53     2438 = 2 * 23 * 53     
1210 = 2 * 5 * 11^2     1331 = 11^3     2662 = 2 * 11^3     
1456 = 2^4 * 7 * 13     1537 = 29 * 53     3074 = 2 * 29 * 53     
1540 = 2^2 * 5 * 7 * 11     1633 = 23 * 71     3266 = 2 * 23 * 71     
1660 = 2^2 * 5 * 83     1837 = 11 * 167     3674 = 2 * 11 * 167     
1780 = 2^2 * 5 * 89     1969 = 11 * 179     3938 = 2 * 11 * 179     
1804 = 2^2 * 11 * 41     1909 = 23 * 83     3818 = 2 * 23 * 83     
1806 = 2 * 3 * 7 * 43     1849 = 43^2     3698 = 2 * 43^2     
1936 = 2^4 * 11^2     2047 = 23 * 89     4094 = 2 * 23 * 89     
2058 = 2 * 3 * 7^3     2401 = 7^4     4802 = 2 * 7^4     
2162 = 2 * 23 * 47     2209 = 47^2     4418 = 2 * 47^2     
2260 = 2^2 * 5 * 113     2497 = 11 * 227     4994 = 2 * 11 * 227

In the examples where the odd $x$ is squarefree, it seems that the prime factors are always $2 \pmod 3.$