edit, Monday, December 5: Pedja's argument can me made to work with the addition of a quadratic reciprocity step...
original: Pedja, you use Dirichlet on the arithmetic progression $12 n + 7,$ which is what you are describing by saying $6k+1$ with $k$ odd. Write $k = 2n+1,$ then $6k+1 = 6 (2n+1) + 1 = 12n + 6 + 1 = 12 n + 7.$
Meanwhile, if you are willing to accept the Cebotarev Density Theorem, asymptotically the two positive quadratic forms $x^2 + 12 y^2$ and $3 x^2 + 4 y^2$ represent the same proportion of primes. The first form represents primes $p \equiv 1 \pmod {12},$ while the second form represents 3 and the primes $q \equiv 7 \pmod {12}.$ I am quoting Theorem 9.12 on page 188 of Primes of the Form $x^2 + n y^2$ by David A. Cox.
EDIT: Actually, your attempt is almost correct, there is an elementary proof of this one. As you write above, take the collection of primes $q_j \equiv 7 \pmod{12}$ that is assumed to be complete, up to some largest we will call $q_r.$ Then, just as you wrote, define
$$ w = 3 + 4 \left( q_1 q_2 \ldots q_r \right)^2 $$
We know that $w \equiv 7 \pmod {12}.$ We also know that $w$ is primitively represented by the quadratic form $3 x^2 + 4 y^2.$ That means that $\gcd(x,y)=1,$ where here $x=1,$ and we are writing $w = 3 x^2 + 4 y^2.$
Now, $w$ is not divisible by 2 or 3. The fact that $\gcd(x,y)=1$ means precisely that $w$ is not divisible by any prime $s \equiv 2 \pmod 3.$ This fact comes under the general heading of quadratic reciprocity. All prime factors of $w$ are, in fact , $1 \pmod 3.$ That is, all prime factors of $w$ are either $1 \pmod {12}$ or $7 \pmod {12}.$ If all prime factors of $w$ were $1 \pmod {12},$ then $w$ itself would also be $1 \pmod {12},$ which it is not.
So, in fact, either $w$ is prime, or $w$ has at least one prime factor, call it $q_{r+1},$ such that $q_{r+1} \equiv 7 \pmod {12}.$ By construction, for any $j \leq r,$ we have $\gcd(q_j,q_{r+1} ) = 1.$ That is, either $w$ itself or $q_{r+1}$ is not in the assumed finite list of primes $7 \pmod {12},$ contradicting the assumption that there is a finite list. $\bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc$
EDIT TOO: Maybe not everyone will know this, so I will give the short proof. Suppose we have some prime $s \equiv 2 \pmod 3,$ with $s \neq 2,$ that divides some $ z = 3 u^2 + 4 v^2.$ Then suppose that $s | z,$ which is also written $ z \equiv 0 \pmod s.$ I claim that, in fact, both $s | u$ and $s | v,$ with the result that $\gcd(u,v)$ is larger than one. PROOF: We have $ 3 u^2 + 4 v^2 \equiv 0 \pmod s.$ Assume that $v \neq 0 \pmod s.$ Then $v$ has a multiplicative inverse $\pmod s.$ So we get
$$ 3 u^2 + 4 v^2 \equiv 0 \pmod s,$$
$$ 3 u^2 \equiv - 4 v^2 \pmod s,$$
$$ \frac{3 u^2}{v^2} \equiv - 4 \pmod s,$$
$$ \frac{ u^2}{v^2} \equiv \frac{- 4}{3} \pmod s,$$
$$ \left(\frac{ u}{v}\right)^2 \equiv \frac{- 4}{3} \pmod s.$$
This says that the right hand side is a quadratic residue $\pmod s.$ It also means that $-12$ is a quadratic residue $\pmod s.$ This is false, though, as we have Legendre symbol
$$( -12 | s) = ( -3 | s) = ( -1 | s ) \cdot ( 3 | s).$$
Now, if $s \equiv 1 \pmod 4,$ we can ignore the $-1$ and switch the other pair. If $s \equiv 3 \pmod 4,$ then $( -1 | s ) = -1$ and $(3 | s) = - (s | 3)$.
In either case, we get
$$ (-3 | s) = (s | 3).$$
As $s \equiv 2 \pmod 3,$ we have
$$ ( s | 3) = ( 2 | 3) = -1.$$
So, in the end we have
$$( -12 | s) = -1,$$
and this contradicts the assumption that $v$ is nonzero $\pmod s.$ Thus, $v \equiv 0 \pmod s.$ From $ 3 u^2 + 4 v^2 \equiv 0 \pmod s,$ we then have $u \equiv 0 \pmod s.$ So $ s | \gcd(u,v)$ and $\gcd(u,v) \neq 1.$
Best Answer
This question has been asked many times. These are called Euclidean proofs of special cases of Dirichlet's theorem on primes in arithmetic progressions. Keith Conrad has a nice article on it, which includes a complete characterization of when such a proof exists. Thanks to the characterization, since $$4^2\equiv 1\pmod{15},$$ there exists such a proof in the case that you requested, though I don't know what explicit polynomial would be used.
I wrote about this general problem in this post as well.
By the way, a Euclidean proof does exist in the $1\pmod 4$ case. You can use the polynomial $n^2+1,$ but you need to include $2$ in the product of the presumed finite list of prime to get the contradiction.
EDIT: I found a polynomial for $15k+4$ with proof in pages 92-94 of Problems in Algebraic Number Theory. It is $$n^4-n^3+2n^2+n+1.$$ The brother of one of the authors recommended the book to me years ago, but it never suited me... finally found use for it.