AustriaMO2013

Since the left side of the equation is certainly larger than 1, we first note that $m>0$ must certainly hold.

Now, we consider even values of $n$. Since $n^2+1\equiv 1(\mathrm{mod}4)$ and $44n^3+11n^2+10n+2\equiv 2(\mathrm{mod}4)$ are certainly true, we have $N^m\equiv 2(\mathrm{mod}4)$. If $m>1$, $N^m$ is odd for any odd $N$ and divisible by 4 for any even $N$, and it follows that $m=1$ must hold, as claimed.

Next, we consider odd values of $n$. In this case we have $44n^3+11n^2+10n+2\equiv 3(\mathrm{mod}4)$ and $n^2+1\equiv 2(\mathrm{mod}4)$, and we see that the factor 2 is contained in $N^m$ exactly $2^k$ times.

For $k=0$ we obtain $N^m=(n^2+1)(44n^3+11n^2+10n+2)\equiv 2(\mathrm{mod}4)$, and the same argument holds as for even values of $n$.

For $k>0$, the exponent $m>1$ must be a divisor of the exponent $k$ of 2 in the prime decomposition of $N^m$, and therefore a power of 2. This means that $(n^2+1{)}^{2k}$ is an $m$-th power, this must also be the case for $44n^3+11n^2+10n+2$, and this number must certainly be a perfect square. This is not possible, however, since we have established that this number is $\equiv 3(\mathrm{mod}4)$, and therefore certainly not a perfect square. This case is therefore not possible, and we see that $m=1$ must hold, as claimed. $\square$