It seems to me that you got derailed by a rather unhelpful discussion in Schwartz. There's a lot going on here which I will briefly outline.
The first thing I want to briefly discus is the relevence of representations of the LORENTZ group vs. representations of the POINCARE group.
Lorentz group: Representations are finite dimensional, given by two half integers $(j_1, j_2)$, and not unitary. Spacetime translation is not included. Representations of this group are representations of local field operators $\hat \phi_i(x)$. Under a Lorentz transformation, the indices of $\hat \phi_i(0$) will get mixed together while the spacetime point $0$ will be fixed. Two simple examples include the spin $0$ representation, where $i$ is just $0$, and the $(1/2, 1/2)$ representation, where we instead use the symbols $\hat A_\mu$ and $\mu = 0... 3$. The fact that these finite dimensional representations (which just juggle around the field indices) are not unitary is totally uninteresting because there isn't a natural inner product of local operators, like there is with states, so who cares that they're not unitary!?
Poincare group: Representations are infinite dimensional, unitary, given (by Wigner's theorem) by either a mass $m^2$ and a half integer spin, or in the $m = 0$ case by a half integer helicity. These correspond to particle states. The fact they they are unitary is important because you want the group action to preserve the natural inner product on states. The fact they are infinite dimensional is not really unexpected, the same thing happens in non relativistic QM! For instance, in the spin $0$ case, single particle states are given by a momentum $| p \rangle$. Under a Poincare transformation,
$$
U(\Lambda, a) | p \rangle = e^{- i p a} | \Lambda p \rangle.
$$
The fact that you need to include a state $| \Lambda p \rangle$ for each Lorentz transformation $\Lambda$, this representation is infinite dimensional because your basis of single particle states is spanned by an infinite number of $p$'s. But this is also the case in non relativistic QM, that you have a basis spanned by an infinite number of momenta, and isn't special to quantum field theory, so it's not really that big a deal. It's true that the Poincare group is non compact because of boosts, but its also non compact because of translations, which is why QM is infinite dimensional (there's an infinite number of positions).
Relationship between the two: You can create single particle states by acting on the vacuum with field operators. In the spin $0$ case a single particle state is given by
$$
| p \rangle = \int d^3 x e^{- i \vec p \cdot \vec x} \hat \phi(\vec x) | 0 \rangle
$$
and in the spin $1$ case by
$$
| p, \varepsilon \rangle = \int d^3 x e^{- i \vec p \cdot \vec x} \varepsilon^\mu \hat A_\mu(\vec x) | 0 \rangle
$$
(modulo maybe some integration measure I forgot.) Here $\varepsilon^\mu$ is a polarization vector satisfying $\varepsilon^\mu p_\mu = 0$. Because single particle states are super positions of states like, say, $\hat A_\mu | 0 \rangle$, the spins of the physical particle states can be found by restricting oneself to the rotation part of the Lorentz group, $U(R)$, and just using the standard angular momentum addition formula, where the irreducible representations of the $j_1 \otimes j_2$ representation is $j_1 +j_2 \oplus j_1 + j_2 - 1 \oplus \ldots \oplus |j_1 - j_2|$. So in the $(1/2, 1/2)$ representation of the Lorentz group, we have $1/2 \otimes 1/2 = 1 \oplus 0$. However, the non physical $0$ representation is projected out by the condition $\varepsilon^\mu p_\mu$, getting rid of the negative norm states, yadda yadda yadda. The point is that $j_1 \otimes j_2$ give the possible particle spins, but some of them will be non physical.
The main point, though, is that representations of the Lorentz group are used for talking about field operators, while representations of the Poincare group are used for talking about particle states. Once you realize this, it's completely reasonable that representations of the Lorentz group are not unitary (why should they be?) and that representations of the Poincare group are not finite dimensional (why should they be?).
Dirac-Spinors in relativistic theory are finite-dimensional representations of the Lorentz-group with peculiarity of being non-unitary.
$\psi^\alpha = \Lambda^\alpha_\beta \psi^\beta$
with finite-dimensional $\Lambda \approx 1+\frac{1}{2}\omega_{\mu \nu}M^{\mu\nu}$ where
$M^{\mu\nu}$ is a matrix belonging to the generators of the Lorentz group
and $\omega_{\mu \nu}$ an antisymmetric matrix containing the 3-dim. angle $\bf{\alpha}$ parameter and the 3-dim.velocity $\bf{v}/c$ parameter.
In non-relativistic theory you need the modulus of the wavefunction $|\psi'|^2= (U\psi)^{\dagger} U\psi = \psi^{\dagger}\psi= |\psi|^2$ being invariant, indeed that can only be achieved with unitary representations, for instance with those of the $SO(3)$-group. However, the corresponding modulus $|\psi|^2=\psi^+ \psi= \bar{\psi}\gamma^0\psi$ in the relativistic theory actually transforms like a 0-component of a 4-vector, so it is not invariant under Lorentz-transformations. So non-unitarity of the Lorentz-transformations is not a problem.
However, if you consider the Poincare'-group, one obtains also infinite-dimensional representations with spinors which transform like this:
$\psi'^{\alpha}(x')= U(\Lambda)\psi^\alpha(x) U(\Lambda^{-1})=D(\Lambda^{-1})^\alpha_\beta\psi^\beta(\Lambda x)$
And they are infinite-dimensional and unitary.
For relativistic Weyl-spinors it is rather similar, the formal expressions in this answer are the same apart from the detail that the size of the matrices and their matrix elements then used has to be adapted. Above all a tentative expression like $|\psi|^2\sim \psi^{\dagger \dot{A}}\psi_A$ (if such an expression made any sense I am not sure of)
formed by Weyl-spinors neither does not need to be invariant, therefore unitarity of the finite Lorentz transformation representation is not required.
For more details the literature has to be consulted. A choice in this respect is for instance the book of Srednicki.
Best Answer
1) Why don't we consider finite dimensional representations of this group?
As you said, we ask (anti)unitarity, so it is impossible to find finite-dimensional representation.
2) Why associate the Lorentz group to fields?
The essence of the answer is what Trimok already said in his comment: the "translational part" of the Poincarè group is already represented by the argument of the field. That is for a general multi-component field you postulate the transformation law, for each element $(a,\Lambda)$ of the Poincarè group, given by
$ \psi'(x') = S(\Lambda) \psi(x)$
where $x' = \Lambda x + a$.
It seems a natural request for the transformation rule of a field, think of the non-relativistic case of the Schroedinger field: you expect that the operator creating a particle in position x with spin m=1 is seen as the operator creating a particle in position x+a with spin m=1 by an observer translated with respect to you. Spin, or any "inner" part of the field, should not be afflicted by translations.
I don't know if there is a more deep or rigorous explanation for this.
So you see that fields are distinguished by $S(\Lambda)$, and so only the Lorentz group is relevant for this purpose.
Note that no general request is asked to $S(\Lambda)$, a part for being a representation (I don't remember if it is allowed to be a projective representation though) of the Lorentz group. In the case of the Dirac field, in order to pin down the explicit form of $S(\Lambda)$, it is made another request, that is it leaves invariant the form of the Dirac equation. In the end it turns out it must not necessarily be unitary.