Solution:

Taking the equation modulo $m+1$ we get $(-1)^b = -1$, so $b$ is odd. Hence we can divide the right-hand side by $m+1$ to get $m^{b-1} - m^{b-2} + \ldots - m + 1$. This has an odd number of terms. If $m$ is odd, then each term is odd and so the total is odd, but $(m+1)^{a-1}$ is even (note that $a > 1$). Contradiction, so $m$ is even.

We have $m^b = (m+1)^a - 1$. Expanding the right-hand side by the binomial theorem, and using $b > 1$, we see that $m$ must divide $a$. So $a$ is even also. Put $a = 2A$, $m = 2M$. We can factorise $(m+1)^a - 1$ as $((m+1)^A + 1)((m+1)^A - 1)$. The two factors have difference $2$, so their gcd divides $2$, but both factors are even, so their gcd is exactly $2$.

If $M = 1$ or a power of $2$, then the smaller factor $3^A - 1$ must be $2$, so $A = 1$ and we have $3^A + 1 = 4$, so $(2M)^b = 8$. Hence $M = 1$ and $b = 3$ and we have the solution $(m, a, b) = (2, 2, 3)$.

If $M$ is not a power of $2$, then $M^b > 2^b$, so we must have the larger factor $2 \cdot M^b$ and the smaller factor $2^{b-1}$. But the larger factor is now $> 2^{b+1}$, so the difference between the factors is at least $3$. $2^{b-1} > 2$. Contradiction.