Problem 2044

If $n = 1$, the only solution is $(m, n) = (1, 1)$. Let $n > 1$.

$n$ is not odd, otherwise since $2n^{\frac{n+1}{2}} + 1 > n^{\frac{n+1}{2}} > n > n-1 > 0$ we get
$$(n^{\frac{n+1}{2}})^2 < (m^3)^2 = n^{n+1} + n - 1 < (n^{\frac{n+1}{2}} + 1)^2$$

and $(m^3)^2$ is strictly between two consecutive squares.

$n \neq 2 \pmod 3$, otherwise since $3n^{\frac{2(n+1)}{3}} + 3n^{\frac{n+1}{3}} + 1 > n^{\frac{2(n+1)}{3}} > n > n-1 > 0$ we get
$$(n^{\frac{n+1}{3}})^3 < (m^2)^3 = n^{n+1} + n - 1 < (n^{\frac{n+1}{3}} + 1)^3$$
and $(m^2)^3$ is strictly between two consecutive cubes.

$n \neq 0 \pmod 3$, otherwise
$$n^{n+1} + n - 1 \equiv -1 \equiv (m^3)^2 \pmod 3$$
Therefore, $n \equiv 4 \pmod 6$. Now
$$m^6 + 3 = n^{n+1} + n + 2 \equiv (-1)^{n+1} + 1 \pmod{n+1}$$
since $n$ is even we get that $m^6 + 3 \equiv 0 \pmod{n+1}$. Suppose that prime number $p$ divides $n+1$. Since $n+1 \equiv 5 \pmod 6$ we get $p > 3$ and since $m^6 \equiv -3 \pmod p$ we see that $-3$ is a quadratic residue modulo $p$. Since $p > 3$ we get $p \equiv 1 \pmod 3$. Since all prime divisors of $n+1$ are 1 modulo 3 we get that $n+1 \equiv 1 \pmod 3$ and therefore $n \equiv 0 \pmod 3$ which contradicts $n \equiv 4 \pmod 6$. The only solution is $(m, n) = (1, 1)$.