We have $5b-3<5a-3<3(2a+1)$, so $5b-3=2(2a+1)$ or $5b-3=2a+1$. In first case we have $b=4a/5+1 \Rightarrow 4a/5+1|a-1$ and so $a=10,b=9$. In second case we have $b=(2a+4)/5 \Rightarrow (2a+4)/5|a-1 \Rightarrow a=3;13$ and so we easy get the answer.

Let $a-1=kb$. In order for divisibility to occur we must have $5b-3\geq 2kb+3$, so $-6\geq b(2k-5)$. So $2k-5<0$ and we have 2 cases. $k=1, 2$. If $k=1$, then $2a+1=2b+3$, so $2b+3\mid (5b-3)+(2b+3)=7b$. Notice that $gcd(5b-3, b)=1 ,3$. If $gcd=1$, then $2b+3=7$, so $(a, b)=(3, 2)$. If $gcd=3$, then $2b+3=21$ ,so $(a, b)=(10, 9)$. If $k=2$, then $2a+1=4b+3$. Similarly $4b+3\mid 9b$ and $gcd(4b+3, b)\leq 3$. Notice that $4b+3$ is always odd, so it can't be divided by $2$. So again 2 cases. If $gcd=1$, then $4b+3=9$ and we get a contradiction. If $gcd=3$, then $4b+3=b, 9 , 27$. It's easy to check that we get 2 contradictions and $(a, b)=(13, 6)$. So only $(a, b)=(3, 2), (10, 9), (13, 6)$ work.