Here is a partial answer to your question. Basically, the methods of Goldston, Graham, Pintz, and Yildirim (see here) have some bearing on your question.
Look at the series
$$\sum_{N < n < 2N} \bigg( \chi_1(n) + \chi_1(n + 2) + \chi_2(n + 2) \bigg)
\bigg( \sum_d \lambda_d \bigg)^2$$
where the $\lambda_d$ are the usual Selberg sieve coefficients or some variant, and $\chi_1$ and $\chi_2$ are the characteristic functions of products of exactly one or two primes. Then I worked out a back of a napkin calculation using Theorems 7-9 in GGPY (possibly I made some mistake), and on the Elliott-Halberstam conjecture (a big assumption!) the above is positive, which gives another proof of Chen's theorem (this one conditional).
One nice feature of the GGPY sieve is that it is compatible with the kind of conditions you describe. For example, let $\chi_2$ be the characteristic function of $E_2$ numbers whose prime factors are both 1 mod 4, or split completely in your favorite Galois extension, or some other appropriate condition. (See the GGPY paper as well as a followup which I wrote.) Then the machinery still works. You just multiply the contribution from the $E_2$ numbers by a fourth. In fact you can do better and only sieve on integers $n$ congruent to 1 mod 4, in which case you only have to divide by a half.
According to my napkin calculation, this is numerically not enough, the series is asymptotically slightly negative, and so this argument fails to prove your assertion, even on EH. So perhaps your question really is "difficult" in the way suggested by the parity problem. But this at least shows you how you can hope to prove statements related to what you asked for.
The major difference is the choice of the Selberg Sieve weights. I strongly recommend reading Maynard's paper. It is well written, and the core ideas are nicely explained.
In what follows, I'll give a brief explanation of what Selberg Sieve weights are, why they appear, and what was chosen differently in Maynard's paper compared to that of Goldston Pintz and Yildirim.
Let $\chi_{\mathcal{P}}(n)$ denote the indicator function for the primes. Given an admissible $k$-tuple $\mathcal{H}=\{h_1<h_2<\cdots<h_k\}$ the goal is to choose a non-negative function $a(n)$ so that
$$\sum_{x<n\leq 2x} \left(\sum_{i=1}^k \chi_{\mathcal{P}}(n)(n+h_i)\right)a(n)\geq \rho \sum_{x<n\leq 2x} a(n)\ \ \ \ \ \ \ \ \ \ (1)$$
for some constant $\rho>1$. If this is the case, then by the non-negativity of $a(n)$, we have that for some $n$ between $x$ and $2x$ $$\left(\sum_{i=1}^k\chi_{\mathcal{P}}(n)(n+h_i)\right)\geq \lceil\rho \rceil,$$ and so there are at least $\lceil\rho \rceil$ primes in an interval of length at most $h_k-h_1$. This would imply Zhangs theorem on bounded gaps between primes, and if we can take $\rho$ arbitrarily large, it would yield the Maynard-Tao theorem. However, choosing a function $a(n)$ for which equation $(1)$ this holds is very difficult. Of course we would like to take $$a(n)=\begin{cases}
1 & \text{when each of }n+h_{i}\text{ are prime}\\
0 & \text{otherwise}
\end{cases} ,$$
as this maximizes the ratio of the two sides, but then we cannot evaluate $\sum_{x<n\leq 2x} a(n)$ as that is equivalent to understanding the original problem. Goldston Pintz and Yildirim use what is known as Selberg sieve weights. They chose $a(n)$ so that it is $1$ when each of $n+h_i$ are prime, and non-negative elsewhere, in the following way:
Let $\lambda_1=1$ and $\lambda_d=0$ when $d$ is large, say $d>R$ where $R<x$ will be chosen to depend on $x$. Then set $$a(n)=\left(\sum_{d|(n+h_1)\cdots(n+h_k)} \lambda_d\right)^2.$$ By choosing $a(n)$ to be the square of the sum, it is automatically non-negative. Next, note that if each of $n+h_i$ is prime, then since they are all greater than $x$, which is greater than $R$, the sum over $\lambda_d$ will consist only of $\lambda_1$. This means that $a(n)=1$ when each of the $n+h_i$ are prime, and hence $a(n)$ satisfies both of the desired properties. Of course, we do not want $a(n)$ to be large when none, or even very few of the $n+h_i$ are prime, so we are left with the difficult optimization problem of choosing $\lambda_d$. The optimization in the classical application of the Selberg sieve to bound from above the number of prime-tuples involves diagonalizing a quadratic form, leading to
$$\lambda_d = \mu(d) \left(\frac{\log(R/d)}{\log R}\right)^k,$$
and so
$$a(n)=\left(\sum_{\begin{array}{c}
d|(n+h_{1})\cdots(n+h_{k})\\
d<R
\end{array}}\mu(d)\left(\frac{\log(R/d)}{\log R}\right)^{k}\right)^{2}.$$
This choice is not so surprising since the related function $\sum_{d|n}\mu(d) \left(\log(n/d)\right)^k$ is supported on the set of integers with $k$ prime factors or less. Goldston Pintz and Yildirim significantly modified this, and used a higher dimensional sieve to obtain their results.
The critical difference in the approach of Maynard and Tao is choosing weights of the form
$$a(n)=\left(\sum_{d_{i}|(n+h_{i})\ \forall i}\lambda_{d_{1}\cdots d_{k}}\right)^{2}.$$ This gives extra flexibility since each $\lambda_{d_1,\dots,d_k}$ is allowed depend on the divisors individually. Using this additional flexibility Maynard and Tao independently established equation $(1)$ for any $\rho>1$, with $k$ depending on $\rho$.
Here are four great resources where you can read more:
- Maynard's paper.
- Tao's blog posts.
- Granville's article on the Maynard-Tao theorem and the work of Zhang.
- Soundararajan's exposition of Goldston Pintz and Yildirim's argument.
Best Answer
Yes, Iwaniec proved in "Almost primes represented by quadratic polynomials" (Invent. math. 47(1978) 171–188), that if $f$ is a quadratic polynomial with $f(0)$ an odd integer, then $f$ contains infinitely many elements of $P_2$.
For an arbitrary polynomial $f$ that is irreducible and doesn't have a fixed prime divisor, one can say that $f$ represents infinitely many elements of $P_n$, where $n=1+\deg f$. This was proven by Buhstab in "A combinatorial strengthening of the Eratosthenian sieve method", Usp. Mat. Nauk 22, no. 3, 199–226.