Team Selection Test for IMO

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.

$n\ne 2(\mathrm{mod}3)$, 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.

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