Find the all the pairs $(n,m) \in \mathbb{N} \times \mathbb{N}$ with the property that $ 2^n+3^m $ is divisible by $23$. Find the all the pairs $(n,m) \in \mathbb{N} \times \mathbb{N}$ with the property that $ 2^n+3^m $ is divisible by $23$.
I'm not really sure how to start this one, but since I found it in a book on group theory, such an approach might also be possible.

As $\displaystyle 2^3\cdot3=24\equiv1\pmod{23},$
$\displaystyle 2^n+3^m\equiv0\pmod{23}\iff 2^{n+3m}\equiv-3^m\cdot2^{3m}\equiv-(3\cdot2^3)^m$
$\displaystyle\implies2^{n+3m}\equiv-1\pmod{23}$
Now as $23$ is prime, $\displaystyle a^2\equiv1,a\equiv\pm1\pmod{23}$ for $(a,23)=1$
and $\displaystyle2^5=32\equiv9,2^{10}\equiv9^2=81\equiv12,2^{11}\equiv2\cdot12\equiv1$
So, there is no integer exponent$(e)$ of $2$ such that $2^e\equiv-1\pmod{23}$
Hence, the system admits no solution in integers

Alternatively,
HINT:
As $3^3\equiv2^2\pmod{23},$
Case $1:$ If $\displaystyle m=3r+1,2^n+3^m=2^n+3^{3r+1}\equiv 2^n+3\cdot2^{2r}\pmod{23}$
So, we need $\displaystyle 2^{n-2r}\equiv-3\pmod{23}$
Observe that there is no integer exponent$(e)$ of $2$ such that $2^e\equiv-3\pmod{23}$
Case $2:$ If $\displaystyle m=3r+2$
Case $3:$ If $\displaystyle m=3r$
