On a (possible?) equivalence of Bunyakovsky conjecture Dirichlet's Theorem on primes in arithmetical progressions is equivalent to the following statement: for all integer numbers $a,b$, where $\gcd(a,b)=1$, there exists at least one prime of the form $an+b$, with $n$ integer. We may call it "a finiteness condition". It is straightforward to prove.
My question is: Does there exists a finiteness condition for Bunyakovsky conjecture? That is, if for all irreducible polynomials $f(x)\in\mathbb{Z}[x]$ with positive leading coefficient and $\gcd(f(1),f(2),f(3),...)=1$ there exist at least a finite number of primes of the form $f(n_1),...,f(n_k)$ for integers $n_1,...,n_k$ (the number $k$ may depend of the polynomial; for linear polynomials, we have $k=1$), is Bunyakovsky conjecture true? I think it is true.
Please note that I am not saying that Bunyakovsky conjecture is true; I just ask if, under the hypothesis given below, it is true.
 A: This is explained in different ways in two places: Chapter 6 of Ribenboim's "Book of Prime Number Records" and an appendix to a paper of Paul Pollack. I'll show how each of them leads to a solution.
First let's write down the conditions that arise in Bunyakovsky's conjecture for a polynomial $f(x) \in \mathbf Z[x]$:


*

*$f(x)$ is nonconstant,

*$f(x)$ is irreducible in $\mathbf Z[x]$,

*for each prime $p$ there is an integer $n$ such that $f(n) \not\equiv 0 \bmod p$.
In the last condition $n$ depends on $p$, and the the last condition is equivalent to saying $\gcd(f(n) : n \in \mathbf Z) = 1.$  Moreover, since $f(n)\bmod p$ only depends on $n$ modulo $p$, the $n$ that is used for a choice of $p$ can be made positive or even sufficiently large. Therefore the last condition can also be described as $\gcd(f(n) : n \in \mathbf Z^+) = 1$ or as $\gcd(f(n) : n \gg 0) = 1$.
We'll call the three conditions above, collectively, the Bunyakovsky conditions. (It's only the third condition that is actually due to Bunyakovsky, but we need all three to have a meaningful conjecture.)
I will now state Bunyakovsky's conjecture in two forms, one using sampling of the polynomial on $\mathbf Z$ and the other using sampling of the polynomial on $\mathbf Z^+$.
Conjecture 1: For every $f(x) \in \mathbf Z[x]$ satisfying the Bunyakovsky conditions, $|f(n)|$ is prime for infinitely many $n \in \mathbf Z$.
Conjecture 1+: For every $f(x) \in \mathbf Z[x]$ satisfying the Bunyakovsky conditions, $|f(n)|$ is prime for infinitely many $n \in \mathbf Z^+$. 
Clearly Conjecture 1+ implies Conjecture 1. I am not aware of an argument in the other direction. Whether you prefer to use the label "Bunyakovsky's Conjecture" to mean Conjecture 1 or Conjecture 1+ is a matter of taste; both are equally inaccessible to proof aside from the linear case. Ribenboim and Pollack each consider "Bunyakovsky's Conjecture" to mean Conjecture  1+. (They also add to the list of Bunyakovsky's conditions that $f(x)$ has a positive leading coefficient and replace $|f(n)|$ being prime with $f(n)$ being prime; this introduces essentially no change in the arguments below, so I leave it out.)
We have given two versions of Bunyakovksy's conjecture.  Here are the seemingly weaker versions that the OP asked about.
Conjecture 2: For every $f(x) \in \mathbf Z[x]$ satisfying the Bunyakovsky conditions, $|f(n)|$ is prime for some $n \in \mathbf Z$.
Conjecture 2+: For every $f(x) \in \mathbf Z[x]$ satisfying the Bunyakovsky conditions, $|f(n)|$ is prime for some $n \in \mathbf Z^+$. 
Trivially Conjecture 1 implies Conjecture 2 and Conjecture 1+ implies Conjecture 2+.  We want to show the converses: Conjecture 2 implies Conjecture 1 and Conjecture 2+ implies Conjecture 1+.  First I'll show Conjecture 2+ implies Conjecture 1+ using the argument given by Ribenboim, and then I'll show Conjecture 2 implies Conjecture 1 using the argument given by Pollack. They each involve a linear change of variables $x \mapsto ax + b$: for Ribenboim $a = 1$ and for Pollack $b = 0$.
Conjecture 2+ implies Conjecture 1+: let $f(x) \in \mathbf Z[x]$ satisfy the Bunyakovsky conditions. By Conjecture 2+, $|f(n_1)|$ is prime for some positive integer $n_1$. Suppose $|f(n_i)|$ is prime for positive integers $n_1 < \cdots < n_r$. We will find a positive integer greater than $n_r$ at which $|f(x)|$ is prime.
Set $g(x) = f(x + n_r)$.  Then $g(x)$ easily satisfies the Bunyakovsky conditions (including the third one), so by Conjecture 2+ for $g(x)$,  there is a positive integer $n$ such that $|g(n)|$ is prime. Therefore $|f(n_{r+1})|$ is prime where $n_{r+1} = n + n_r > n_r$.
(Edit: This argument that Conjecture 2+ implies Conjecture 1+, which I saw in Ribenboim's book, is taken from the paper of Schinzel and Sierpinski about Schinzel's Hypothesis H, which is the multi-polynomial generalization of Bunyakovsky's conjecture; see p. 188 of the paper here.)
Conjecture 2 implies Conjecture 1: let $f(x) \in \mathbf Z[x]$ satisfy the Bunyakovsky conditions. Since $f(x)$ is nonconstant, it has a composite value $f(c)$ at some positive integer $c$ (see a proof here). Replacing $f(x)$ with $f(x+c)$ does not change the Bunyakovsky conditions being satisfied or change the conclusion of Conjecture 1 or 2 being satisfied, so without loss of generality $f(0)$ is composite. 
By Conjecture 2, $|f(n_1)|$ is prime for some integer $n_1$. Suppose $|f(n_i)|$ is prime for distinct integers $n_1, \ldots, n_r$. We will find another another integer $n_{r+1}$ such that $|f(n_{r+1})|$ is prime.
Since there are infinitely many primes, we can pick a prime number $q$  greater than $\max(|n_1|,\ldots,|n_r|)$ that does not divide $f(0)$. Set $F(x) = f(qx)$. Let's check it satisfies the Bunyakovsky conditions:


*

*Clearly $F(x)$ is nonconstant.

*Being irreducible in $\mathbf Z[x]$ is equivalent to being primitive (coefficients have gcd 1) and being irreducible in $\mathbf Q[x]$. Clearly $F(x)$ is irreducible in $\mathbf Q[x]$, and it is primitive since $f(x)$ is primitive and $q$ does not divide the constant term $F(0) = f(0)$.

*If the third Bunyakovsky condition were false for $F(x)$ then there is a prime $\ell$ such that $F(n) \equiv 0 \bmod \ell$ for all integers $n$. Write that as $f(qn) \equiv 0 \bmod \ell$ for all integers $n$. If $\ell \not= q$ then $q$ is invertible mod $\ell$, so $f(m) \equiv 0 \bmod \ell$ for all integers $m$, which contradicts the third Bunyakovsky condition for $f(x)$. If $\ell = q$ then $f(0) \equiv 0 \bmod q$, but $q$ was chosen not to divide $f(0)$. Thus there is no such $\ell$, so $F(x)$ satisfies the third Bunyakovsky condition. 
By Conjecture 2 for $F(x)$, there is an integer $n$ such that $|F(n)|$ is prime, so $|f(qn)|$ is prime. Since $F(0) = f(0)$ is not prime, $n$ is not $0$. Thus $|qn| \geq q > \max(|n_1|,\ldots,|n_r|)$, so we can use $qn$ as $n_{r+1}$. 
In both of these proofs, we apply whichever weak form of Bunyakovsky's conjecture we assume to both $f(x)$ and either $f(x+k)$ or $f(kx)$ for nonzero integers $k$.  Therefore we could have fixed the degree of the polynomials in Conjectures 1 and 2 or Conjectures 1+ and 2+ and still gotten an equivalence. The case of degree $1$ is precisely the straightforward equivalence mentioned in the original question.
UPDATE: There is an extension of Bunyakovsky's conjecture to simultaneous primality of multiple polynomials. For nonconstant $f_1(x), \ldots, f_r(x)$ in $\mathbf Z[x]$, it tells us when to expect $|f_1(n)|, \ldots, |f_r(n)|$ are all prime for infinitely many integers $n$, or for infinitely many positive integers $n$. The analogue of the Bunyakovsky conditions is the following: 


*

*$f_1(x), \ldots, f_r(x)$ are all nonconstant,

*$f_1(x), \ldots, f_r(x)$ are each irreducible in $\mathbf Z[x]$, 

*for each prime $p$ there is an integer $n$ such that $f_1(n)\cdots f_r(n) \not\equiv 0 \bmod p$.
This third condition uses the product of the $f_i(x)$ rather than the individual $f_i(x)$.  For example, there are not infinitely many prime pairs $n, n+1$ since one of them is always even, and that violates the third condition at the prime $2$: $n(n+1) \equiv 0 \bmod 2$ for all integers $n$.
A set of $r$ nonconstant polynomials in $\mathbf Z[x]$ whose absolute values are all prime simultaneously at infinitely many integers (or positive integers) necessarily satisfies the 2nd and 3rd conditions above.  That these conditions should be sufficient is called 
Schinzel's Hypothesis H. When $r = 1$ this is Bunyakovsky's conjecture. 
The reasoning used above to equate Conjectures 1 and 1+, and Conjectures 2 and 2+, can be extended from one polynomial to multiple polynomials:
if for every $r$-tuple $f_1(x), \ldots, f_r(x)$ in $\mathbf Z[x]$ fitting the three conditions above there is some $n \in \mathbf Z^+$ such that all $|f_j(n)|$ are prime, then for every $r$-tuple $f_1(x), \ldots, f_r(x)$ in $\mathbf Z[x]$ fitting the three conditions above there are infinitely many $n \in \mathbf Z^+$ such that all $|f_j(n)|$ are prime, and similarly if we replace "$n \in \mathbf Z^+$" by "$n \in \mathbf Z$" in both places. The only new thing you have to check in order to generalize the reasoning from one polynomial to $r$ polynomials is that for every list of $r$ nonconstant polynomials $f_1(x), \ldots, f_r(x)$ in $\mathbf Z[x]$ there is a positive integer $c$ (in fact, there are infinitely many) at which all $f_i(c)$ are composite numbers.  Further details are left to the reader.
