The Lagrangian density of a Poincare invariant theory should not depend explicitly on the space-time coordinates. Does this mean $$ \partial_\mu \mathcal{L}=0~? $$
If this is the case doesn't the above equation imply that $\mathcal{L}$ is a constant? I know that $\mathcal{L}$ is a functional not a function but at the end it is still a function of space-time.
I) OP wrote (v2):
I know that $\mathcal{L}$ is a functional not a function.
Actually for local theories the Lagrangian density $$ \mathcal{L}(\phi(x), \partial\phi(x), \partial^2\phi(x), \ldots, ;x) $$ is a function of $\phi(x)$, $\partial\phi(x)$, $\partial^2\phi(x)$, $\ldots$, and $x$. In contrast, the action $S[\phi]=\int \!d^dx~\mathcal{L}$ is a functional of $\phi$, cf. e.g. this Phys.SE post.
II) OP wrote (v2):
The Lagrangian density of a Poincare invariant theory should not depend explicitly on the space-time coordinates. Does this mean $$\tag{1} \partial_{\mu} \mathcal{L}~=~0~? $$
If $\partial_\mu$ in OP's notation denotes explicit space-time differentiation $ \frac{\partial}{\partial x^{\mu}}$, then the answer is Yes. Note however that there are also implicit dependence on the space-time coordinates via the fields $\phi$. The total space-time derivative becomes
$$ d_{\mu}\mathcal{L}~=~\partial_{\mu} \mathcal{L} + \frac{\partial\mathcal{L}}{\partial\phi}\partial_{\mu}\phi + \frac{\partial\mathcal{L}}{\partial(\partial_{\nu}\phi)} \partial_{\nu}\partial_{\mu}\phi + \ldots . $$
III) Translational invariance (1) implies via Noether's theorem that the corresponding energy-momentum 4-vector $P_{\mu}$ is conserved.
Poincare invariance simply means that the Lagrangian does not change under a Poincare transformation:
$$\mathcal{L}(x)\rightarrow\mathcal{L'}(x)=\mathcal{L}(\Lambda^{-1}x).$$
Note that I have only written down Lorentz transformations, for simplicity. We now say that $\mathcal{L}$ is invariant if
$$\mathcal{L}(x)=\mathcal{L}'(x).$$
This does not mean that the Lagrangian does not depend on spacetime coordinates. After all, the fields it contains are supposed to be functions of spacetime. It just tells us something about the transformation behaviour of the function. An invariant Lagrangian can be constructed out of expressions that transform like scalars, i.e. $\phi(x)\rightarrow\phi(\Lambda^{-1}x).$
You could say that $\partial_{\mu}\mathcal{L}=0$ only if the fields it applies to are treated as independent variables and not as functions of $x^\mu$. I.e. the Lagrangian does not explicitly refer to coordinates, but if you were to consider the fields as $\phi = \phi(x^\mu)$, then obviously $\partial_{\mu}\mathcal{L}\neq 0$.
The more rigorous way to say this is by considering a transformation which translates all the relevant physics in space-time by $a^\mu$. The Lagrangian gets transformed to $\mathcal{L}(x^\mu) \to \mathcal{L}_a(x^\mu)$ - for now we do not assume any form of this transformation. The zero derivative requirement could be specified as $$\mathcal{D}_a \mathcal{L} \equiv \lim_{\epsilon \to 0} \frac{\mathcal{L}_{a \epsilon}(x^\mu + \epsilon a ^\mu)-\mathcal{L}(x^\mu)}{\epsilon} = 0$$ I.e. if you do an infinitesimal translation of all the physics and compare the result with the original untranslated one (you must compare at the back-translated point), you get the same result. But by requiring this and assuming some reasonable mathematics, we get that it is equivalent to the condition $$ \mathcal{L}_a (x^\mu) = \mathcal{L}(x^\mu-a^\mu)$$ I.e. translating physics and back-translating coordinates yields the same result even if not infinitesimal. The important part is obviously the "all relevant physics" translating with this $a$-translation. This is almost a tautological statement - the Lagrangian depends only on "all relevant physics" and it's value gets transported with them.
The latter condition is a more special case of the ones mentioned by Frederic Brünner and m1rohit.
