Are there infinitely many pairs of positive integers $(a,b)$ such that $2(6a+1)$ divides $6b^2+6ab+b-6a^2-2a-3$? That is, if there are infinitely many different $a$ and for which at least one value of $b$ can be found for a given $a$. Some $a$ values are $0,2,3,7,11,17....$ I think that the answer is yes but I have no idea how to show this.
If the answer is yes, are there any polynomial parametric solutions?
If $6a+1$ is a prime and a quadratic residue modulo $17$ (which is true for infinitely many values of $a$), then there are infinitely many positive integers $b$ with the required property.
First observe that $b$ is good if and only if $$f(a,b):=6b^2+6ab+b-6a^2-2a-3$$ is even and divisible by $6a+1$. Hence $b$ must be odd: $b=2c+1$. Now we need to find infinitely many positive integers $c$ such that $f(a,2c+1)$ is divisible by $6a+1$. The identity $$6f(a,2c+1) + (6a-5-12c)(6a + 1)=(12c+6)^2-17$$ shows that $c$ is good if and only if $(12c+6)^2-17$ is divisible by $6a+1$. So we need to guarantee that the congruence $$(12c+6)^2\equiv 17\pmod{6a+1}$$ has a solution. This is equivalent to $17$ being a quadratic residue modulo $6a+1$. By quadratic reciprocity, this is equivalent to $6a+1$ being a quadratic residue modulo $17$, and we are done.
