First, let's determine how large $log_a b$ is. Note that $log_a b > 0$ since $b > 1$ and $a\ge  2$ while $b\le 2015$ so $log_a b \le log_2 2015 < 11$.

• If $log_a b = 1$, then $a = b$ so there are $2014$ such pairs.
• If $log_a b = 2$, then $a^2 = b \le 2015$ so $a < 45$ so there are $43$ such pairs.
• If $log_a b = 3$, then $a^3 = b\le 2015$ so $a < 13$ so there are $11$ such pairs. Note that we can find this bound easily by $12^3 = 144\times 12 < 150 \times 12 = 1800$ and $13^3 = 169\times 13 > 160\times 13 = 2080$.
• $\cdots$

We continue for each integer value of loga b less than 11 and in total there are $2014+43+11+5 + 3 + 2 + 1 + 1 + 1 + 1 = 2082$ such pairs.