For brevity, we name $n = a^b + b$ and $m = a^{2b} + 2b$.

For $a = b = 0$, we obtain $n = m = 1$, and therefore $n|m$. We see that $(0, 0)$ is a solution.

For $a = 0$ and $b > 0$, we obtain $n = b$ and $m = 2b$, and again $n|m$. We see that $(0, b)$ is, in fact a solution for all values of $b$.

For $a = 1$, we obtain $n = b + 1$ and $m = 2b + 1$. The only way $(b+1)|(2b+1)$ can hold is for $b = 0$. We see that $(1,0)$ is also a solution.

For $b = 0$, we obtain $n = 1$ and $m = 1$, and again $n|m$. $(a,0)$ is therefore a solution for all values of $a$.

We now consider the general case, where $a > 1$ and $b > 0$. Since
$$a^b + b \mid a^{2b} + 2b = (a^b)^2 - b^2 + (b^2 + 2b) = (a^b + b)(a^b - b) + b(b+2),$$
we see that $a^b + b \mid b(b+2)$ must hold. For $b=1$, we must have $a+1|3$, and therefore $a=2$. The only solution in this case is therefore $(2, 1)$.

We now consider large values of $a$, specifically $a \ge 3$. For $a=3$, we have $n = 3^b + b$, and we can show that $3^b + b > b(b+2) \iff 3^b > b(b+1)$, which yields a contradiction. We show this by induction. For $b=1$, we have $3 > 1 \cdot 2$, which is certainly true. If we can now assume that $3^k > k(k+1)$ is true, and wish to show $3^{k+1} > (k+1)((k+1)+1)$, we note that the latter expression results from the former by multiplying the left side with $3$ and the right side with $\frac{k+2}{k}$, which is certainly not greater than $3$, which proves the induction. If $a > 3$, we have $n > 3^b + b$, which is then also certainly impossible.

Finally, the case $a=2$ remains. As before, we can show that $2^b > b(b+1)$ for $b > 5$ (noting $32 > 30$). The only possible values for $b$ remaining are therefore $2$, $3$ and $4$. For $b=2$, we would have $6|8$, for $b=3$ we would have $11|21$, and for $b=4$ we would have $20|24$, none of which is true.

In summary, the solutions are $(2, 1)$, all pairs $(a, 0)$ and all pairs $(0, b)$. qed