Noether's current for dilation transformation Consider the Lagrangian of $\phi^4$ theory
$$
\mathcal{L} = \frac{1}{2}\partial_\mu\phi \partial^\mu\phi - \frac{\lambda}{4!}\phi^4.
$$
We define the following dilation transformation
$$
x^\mu \rightarrow e^\alpha x^\mu,\\
\phi(x) \rightarrow e^{-\alpha}\phi(x).
$$
How do we find the Noether current associated with this transformation? We can see that the action $S = \int d^4x \mathcal{L}$ is invariant under dilation because
$$
d^4x \rightarrow e^{4\alpha} d^4x,\\
\partial_\mu \rightarrow e^{-\alpha} \partial_\mu,\\
\mathcal{L} \rightarrow e^{-4\alpha} \mathcal{L}.
$$
The problem with $\mathcal{L} \rightarrow e^{-4\alpha} \mathcal{L} = \mathcal{L} -4\alpha\mathcal{L} + O(\alpha^2)$ is that it is not of the form $\mathcal{L} \rightarrow \mathcal{L} + \alpha \partial_\mu J^\mu$, so how do we find the conserved current?
Edit: Sorry $m$ is supposed to be $0$
 A: Firstly is there a transformation on $m^2$? Otherwise what happens to the second term? 
Expand the variation of the Lagrangian to first order in $\alpha$ - Noether's theorem uses the infinitesimal form of the transformation. Then you can look for the solution to the equation  $\partial _\mu J^\mu =-\mathcal{L} $.
Substantial edit:
In order to examine the conserved current one needs to determine the infinitesimal transformations of the fields. Let's work them out. Firstly the finite transformations are
$$\begin{align}
x &\longrightarrow x' = \textrm{e}^{\alpha}x \\
\phi(x) & \longrightarrow \phi'(x') = \textrm{e}^{-\alpha}\phi(x)
\end{align}$$
where the second line follows from the definition of $\phi$ as a scalar field of conformal weight given in OP's question. Now we define infinitesimal variations in the field as follows:
$$ \delta_{\alpha}\phi(x) = \phi'(x) - \phi(x)$$
and we'll be interested in the variation to linear order in $\alpha$. 
For us, we will make an active transformation so that with $\phi'(y') = \phi'(\textrm{e}^{\alpha}y) = \textrm{e}^{-\alpha}\phi(y)$ we deduce that $\phi'(x) = \textrm{e}^{-\alpha}\phi(\textrm{e}^{-\alpha}x)$. To find the Noether current we should expand to linear order in $\alpha$ which leads to
$$\phi'(x) = (1 - \alpha)\phi(x - \alpha x) + \mathcal{O}(\alpha)= \phi(x) - \alpha\left(1 + x \cdot \partial\right)\phi(x)+ \mathcal{O}(\alpha)$$
and we have found
$$\delta_{\alpha}\phi(x) = -\alpha\left(1 + x \cdot \partial\right)\phi(x)$$
Now we will continue to find the variation in the derivative of $\phi
$:
$$\delta_{\alpha}\partial_{\mu}\phi(x) = -\alpha\left(2 + x \cdot \partial\right)\partial_{\mu}\phi(x)$$
where I assumed that $\alpha$ is constant. 
Now we recall that construction of the current: if $\delta_{\alpha}\mathcal{L} = \alpha \partial_{\mu}f^{\mu}$ then the current defined by $\alpha J^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\delta_{\alpha}\phi - \alpha f^{\mu}$ satisfies the continuity equation $\partial_{\mu}J^{\mu} = 0$. 
In the current case, then I leave it as an exercise to verify 
$$\delta_{\alpha} \mathcal{L} = -\alpha \partial_{\mu}\left(x^{\mu}\mathcal{L}\right)$$
so that $f^{\mu} = -x^{\mu}\mathcal{L}$ and 
$$-J^{\mu} = \phi \partial^{\mu}\phi + (\partial^{\mu}\phi) (x \cdot \partial \phi) - x^{\mu}\left(\frac{1}{2}\partial_{\nu}\phi \partial^{\nu}\phi - \frac{\lambda}{4!}\phi^{4}\right)$$
and to check that the equations of motion imply that $\partial_{\mu}J^{\mu} = 0$. 
A: There is a general formulae for the conserved current due to such translation symmetries. Indeed, assume that under the symmetry $x\mapsto x'=x+\alpha x$ and $\phi\mapsto \phi'$, with $\phi'(x')=\phi(x)+\alpha d\phi(x)$ and $\alpha$ is infinitesimal, the action is invariant. More precisely if $\Omega$ is a region of spacetime and $\Omega'$ is the dilation of this region, we assumet that $S_{\Omega'}(\phi')=S_\Omega(\phi)$. This fixes the transformation behavior of the Lagrangian. Indeed,
$$S_{\Omega'}(\phi')-S_\Omega(\phi)=\int_\Omega \left(\delta(d^Dx)\mathcal{L}+d^Dx\bar{\delta}\mathcal{L}\right).$$
In here $\bar\delta$ denotes the variation $\bar{\delta}F(x)=F'(x')-F(x)$. In terms of the functional variation $\delta F(x)=F'(x)-F(x)$ we have $\bar{\delta}=\delta x^\mu\partial_\mu+\delta$. For example $\delta\phi=\alpha d\phi-\alpha x^\mu\partial_\mu\phi$. Using $\bar{\delta}\mathcal{L}=\delta\mathcal{L}+\delta x^\mu\partial_\mu\mathcal{L}$ and $\delta(d^D x)=d^D x\partial_\mu\delta x^\mu$, we obtain
$$0=S_{\Omega'}(\phi')-S_\Omega(\phi)=\int_\Omega d^Dx \left(\partial_\mu(\delta x^\mu\mathcal{L})+\delta\mathcal{L}\right).$$
We conclude that $\delta\mathcal{L}=-\alpha\partial_\mu(x^\mu\mathcal{L})=\partial_\mu F^\mu$ and thus we have the conserved current
$$j^\mu=\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\delta\phi-F^\mu=\alpha\left(\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}(d\phi-x^\nu\partial_\nu\phi)+x^\mu\mathcal{L}\right).$$
One can specialize this result to your case by taking $d=-1$.
