Note that if $\nu_p(a)\neq\nu_p(b)$, then $\nu_p(a+b)=\min(\nu_p(a), \nu_p(b))$. Suppose $b$ has the property and $p$ is a prime divisor of $b$. If $k\nu_p(a)=\nu_p(a^k)=\nu_p(b^l)=l\nu_p(b)$, then $\nu_p(a^l)=l\nu_p(a)=\frac{l^2}k\nu_p(b)\neq k\nu_p(b)=\nu_p(b^k)$ since $k^2\neq l^2$. Therefore, either $\nu_p(a^k)\neq\nu_p(b^l)$ or $\nu_p(a^l)\neq\nu_p(b^k)$. $\therefore\min(\nu_p(a^k+b^l), \nu_p(a^l+b^k))\leq\max(\min(\nu_p(a^k), \nu_p(b^l)), \min(\nu_p(a^l), \nu_p(b^k)))\leq\max(\nu_p(b^l), \nu_p(b^k))<\nu_p(b^{k+l})$. $\Rightarrow b^{k+l}$ cannot divide both $a^k+b^l$ and $a^l+b^k$. $\therefore$ the only possible $b$ is $1$.

Another solution: W.l.o.g. $k>\ell$ so that $k=\ell+d$. Then \[b^{k+\ell} \mid a^d(a^{\ell}+b^{\ell+d})-(a^{\ell+d}+b^{\ell})=a^db^{\ell+d}-b^{\ell}.\]Hence $b^k \mid a^db^d-1$ and in particular $b \mid 1$ and hence $b=1$.