Here is a constructive bound for the number of solutions to $\phi(x)=m$. 

Let $\varphi(a) = m$. If $p^k\mid a$ for some $k\geq 1$, then $p^{k-1}(p-1)\mid m$, and thus $(p-1)^k \leq p^{k-1}(p-1)\leq m$, implying that $k\leq \frac{\log(m)}{\log(p-1)}$. Then the number of such $a$ is bounded by

$$\prod_{d\mid m\atop d\geq 2} (1+\frac{\log(m)}{\log(d)})< \exp\left(\sum_{d\mid m\atop d\geq 2} \frac{\log(m)}{\log(d)}\right) = m^E,$$
where
$$E:=\sum_{d\mid m\atop d\geq 2} \frac{1}{\log(d)} = \frac{1}{\log(2)} + \int_{2}^{m} \frac{\mathrm{d}t}{\log(t)}=\frac{1}{\log(2)} + \mathrm{Ei}(\log(m))-\mathrm{Ei}(\log(2)) < 1 + \gamma + \frac{11}{36}\log(m)^2,$$
$\mathrm{Ei}(x)$ is the [exponential integral](https://en.wikipedia.org/wiki/Exponential_integral) and $\gamma$ is [Euler-Mascheroni constant](https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant).

So, we can generously bound the number of solutions to $\phi(x)=m$ by
$$m^{1+\gamma}e^{\frac{11}{36}\log(m)^3}.$$