A generalization of Green-Tao for polynomials of arbitrary degree

Question

The Green-Tao theorem asserts that for any positive integer $M$, there exists an irreducible $p \in \mathbb{Z}[x]$ with $\text{deg}(p) = 1$ such that $p(i)$ for each $0 \leq i \leq M$ is a prime.

Does the following generalization hold?

For any $(d,M) \in \mathbb{N}^2$, there exists an irreducible $p \in \mathbb{Z}[x]$ with $\text{deg}(p)=d$ such that $p(i)$ is prime for each $0 \leq i \leq M$.

My suspicion is that either this is somehow a direct corollary of Green-Tao, or an unsolved problem entirely, the latter seeming more likely. If this is the case, a reference would be great.

It's somewhat related to the well-known unsolved Bunyakovsky conjecture, although here we don't require $p$ to produce infinitely many primes, and the primes need to be consecutive.

What does seem to be known is, for the case $d=2$, there are only finitely many $q$ such that $n^2+n+q$ is prime for $n \in \{0,1,...,q-2\}$, but this doesn't quite rule out polynomials of the form $an^2+bn+q$.

Answer 1

It looks like this 2008 paper by Tao and Ziegler gives you what you want. The abstract begins:

We establish the existence of infinitely many polynomial progressions in the primes; more precisely, given any integer-valued polynomials $P_1, \dots, P_k ∈ \mathbb{Z}[m]$ in one unknown $m$ with $P_1(0) = \dots = P_k(0) = 0$, and given any $\varepsilon > 0$, we show that there are infinitely many integers $x$ and $m$, with $1\le m\le x^\varepsilon$, such that $x + P_1(m), \dots, x + P_k(m)$ are simultaneously prime. The arguments are based on those in [18] [the Green-Tao theorem paper], which treated the linear case $P_j = (j − 1)m$ and $\varepsilon = 1$...

In particular, let $P_j(m)=P(jm)$ for some fixed polynomial $P$. Then this gives us some $x$ and $m$ such that $x+P(m),x+P(2m),\dots,x+P(km)$ are all prime. Setting $p(n)=x+P(mn)$ gives a polynomial of the same degree as $P$ which is prime at the $k$ consecutive integers $n=1,2,\dots,k$.

(Moreover, if $k > 4d$, then $p(n)$ must be irreducible. To see this, suppose that $p(n)=q(n)r(n)$. Then either $q(n)$ or $r(n)$ is equal to $\pm 1$ at each number $1,2,\dots,k$. So by the pigeonhole principle at least one of $q$ or $r$ assumes at least one of the values $\pm 1$ at least $d+1$ times. But the degrees of $q$ and $r$ are at most $d$, so this means that at least one of $q$ and $r$ is a constant polynomial $\pm 1$.)

Note that this is a 90-page paper that extends the proof of the Green-Tao theorem, so not exactly a direct corollary. However, the result it proves is also a lot more general than what you're asking for.