If $6a+1$ is a prime number, 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 $6a+1$. As $6a+1$ is prime, the last property is equivalent to $$12c+6\equiv 17^{3a}\pmod{6a+1}.$$ Equivalently, $$2c+1\equiv -a 17^{3a}\pmod{6a+1}.$$ This clearly has infinitely many positive integer solutions $c$, and we are done.

To sum up, if $6a+1$ is a prime number, then any odd $b$ congruent to $-a 17^{3a}$ modulo $6a+1$ will have the required property.

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.