Prove that $23$ does not divide $2^n + 3^m$, for any $m, n \in \mathbb{N}$ Prove that $23$ does not divide $2^n + 3^m$, for any $m, n \in \mathbb{N}$.

I've tried to prove that the equation $2^n + 3^m = 0$ has no solutions in $\mathbb{Z}_{23}$, but didn't succed.

Thank you!
Hint:  Render $2\equiv 5^2$, $3\equiv 7^2$ mod $23$.  Can a sum of nonzero squares have zero residue mod $23$ when $23$ is a $4k+3$ prime?
Another approach, if the above is &quot;slightly overlevel&quot; as in the comments:
If $2^n+3^m$ were divisible by $23$ then so is $8^n+27^m$ as $a+b$ divides $a^3+b^3$.  Then $8^n+4^m=2^{3n}+2^{2m}$ would also be a multiple of $23$.  Through division by the smaller power of $2$ in $\{8^n,4^m\}$ we find that $2^k+1$ should be a multiple of $23$ for some whole number $k(=|3n-2m|)$.  Now try to find a value of $k$ that makes $2^k\equiv-1\bmod 23$.
There are only 11 modului that $2^n$ and $3^n$ can be, and the other remaining 11 modulii are the 11 numbers required to add be added to give 23.

The fact that 23 is 3, mod 4, but is -1, mod 24, means that if $2^m+3^n$ gives a value, mod 23, then the negative is not in that this form,

The same applies to any prime that gives 23 mod 24, such as 47, 71, 167, ...