Taking equation mod $\mathrm{m}+1$ we get $(-1)^{\mathrm{b}}=-1$, so $\mathrm{b}$ is odd. Hence we can divide the rhs by $\mathrm{m}+1$ 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 $(\mathrm{m}+1)^{\mathrm{a}-1}$ is even (note that $\left.\mathrm{a}>1\right)$. Contradiction, so $\mathrm{m}$ is even.

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

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

If $\mathrm{M}$ is not a power of 2, then $\mathrm{M}^{\mathrm{b}}>2^{\mathrm{b}}$, so we must have the larger factor $2 \cdot \mathrm{M}^{\mathrm{b}}$ 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. $2^{\mathrm{b}-1}>2$. Contradiction.