So $3a+1 \mid 4b-1$ and $2b+1 \mid 3a-1$. The main idea is that the first implies (essentially) $a \le \frac{4b}{3}$ and the first $a \ge \frac{2b}{3}$ so we only need to win by a factor of $2$ (essentially). This motivates the following argument: First case: $2b+1=3a-1$. Then $4b-1=6a-5$, hence $3a+1 \mid 6a-5$, hence $3a+1 \mid 7$, hence $a=2$ and $b=2$ which is a solution. Second case: $3a+1=4b-1$. Then $3a-1=4b-3$, hence $2b+1 \mid 4b-3$, hence $2b+1 \mid 5$, hence $b=2$ and $a=2$ which is the same solution again. Third case: $4b-1 \ge 6a+2$ and $3a-1 \ge 4b+2$. But then $3a-1 \ge 6a+2$, contradiction. Hence the only solution is $(2,2)$.