Almost-all Goldbach for the quadratic sequence $9n^2+1$?

Let $$B:=\{n\geq 13: n \text{ odd and there exists a prime } q \text{ such that } 9n^2+1-q \text{ is prime, where either } q=3, \text{or } q\geq 11,q \equiv 2\pmod 3, q-1 \text{ is not a square}\}$$

I am trying to understand whether the following “almost-all Goldbach” statement is known / correct:

For almost all odd integers $(n\ge 13)$, the even number $9n^2+1$ can be written as a sum of two primes.

Equivalently, if $O:=\{n\ge 13:\ n\ \text{odd}\}$ is it true that $d(B)=\frac{1}{2}$ or, in other words, $d(O\setminus B)=0$?

Here $d(B)$ means $d(B) := \lim_{N \to \infty} \frac{|B \cap \{1,2,3,...,N\}|}{N}$, where $B$ is a set.

My reasoning is the following.

For odd $n$, write $n=2t+1$. Then $9n^2+1 = 9(2t+1)^2+1 = 2(18t^2+18t+5). $

So this is a quadratic polynomial sequence of even numbers.

I know there are “almost all” results for Goldbach representations along polynomial sequences, and it seems plausible that such a theorem should imply that for almost all odd $n$, $9n^2+1=p+q$ with primes $p$, and $q$.

If that is true, then there is one extra issue in my definition of $B$: I exclude the case where $q-1$ is a perfect square. My hope is that this exclusion is harmless, because primes of the form $q=a^2+1$ are very sparse: up to $X$, there are at most $\sqrt X$ such primes.

So my questions are:

1. Is it known that almost all numbers of the form $9n^2+1$ (with $n$ odd) satisfy Goldbach?

2. If yes, is it legitimate to conclude that the additional restriction

   $$q-1\ \text{is not a perfect square}$$

   only removes a density-zero subset, so that $d(B)=1/2$?

3. Is there a standard reference that directly covers this special case?

I would also be happy with a reference showing only that

$$|\{n\leq x: n \text{ odd },\ 9n^2+1 \text{ is a sum of two primes}\}| \leq \frac{x}{2}+o(x)$$

or even just a weaker asymptotic lower bound implying infinitely many such $n$.

Thanks.

A second version can be made more concise and more reference-focused if you want.