A "general rule" is that problems like this simplify when using CRT to split them into congruences modulo smaller moduli. Doing that here reduces to simple parity arguments, namely
$\qquad\begin{align} &\bmod 20\!:\,\ 3^{\large 1+2j}+ 2^{\large 1+2k}\equiv 15^{\large z}\\[.2em]
\iff &\bmod 20\!:\,\ 3\cdot 9^{\large j}\ + 2\cdot 4^{\large k}\ \equiv 15^{\large z}\\[.2em]
\Rightarrow\ &\bmod 4\!:\,\ {-}1\equiv (-1)^{\large z} \!\iff z\equiv 1\!\!\pmod{\!2}\\[.2em]
\Rightarrow\ &\bmod 5\!:\,\ 3(-1)^{\large j}-3(-1)^{\large k}\equiv 0\iff (-1)^{\large j}\equiv (-1)^{\large k}\iff j\equiv k\!\!\pmod{\!2}
\end{align}$