Divisibility and number theory in terms of a and b

Sure, choose for example $a=2k-1,b=2k+1$ for any integer $k$. Clearly these pairs are coprime, and we have

\begin{align} a^b+b^a&=(2k-1)^{2k+1}+(2k+1)^{2k-1}\\ &\equiv -1+(2k+1)2k+1+(2k-1)2k\tag{*}\\ &=8k^2\\ &\equiv 0 \pmod{4k}. \end{align}

where $(*)$ follows from Binomial theorem for example (or see jgon's answer for alternative argument). So $a^b+b^a \equiv 0 \bmod {a+b}$ since $a+b=4k$.

Sil posted a now deleted almost correct answer, so this is based on those ideas. (The answer has since been undeleted and corrected)

Let $a=2k-1$, $b=2k+1$ for any integer $k$. $a,b$ are coprime. $a+b=4k$.

Note that $$ab=4k^2-1 \equiv -1 \pmod{4k},$$ $$a^2 = 4k^2 - 4k +1 \equiv 1 \pmod {4k},$$ $$b^2 = 4k^2 + 4k +1 \equiv 1 \pmod {4k}.$$ Since $ab\equiv -1\pmod{4k}$, $b^{-1} \equiv -a\pmod{4k}$.

Then $$ a^b+b^a =a^{2k+1}+b^{2k-1} \equiv a^1 + b^{-1} \equiv a-a =0\pmod{4k}. $$

Thus $a^b+b^a \equiv 0\pmod{a+b}$, so $a+b\mid a^b+b^a$.

Hence we have found infinitely many such pairs.