This is an entry to the 12th fortnightly challenge.
This maze is built out of the integers $1,\dots,560$. You may step from $k$ to $2k$, or from $3k+1$ to $5k+1$, or vice versa in each case, provided you don't step outside that range $1,\dots,560$. ($k$ is an integer.)
So here is a small part of the maze: $$ \begin{array}{c} &&&&21& \longleftrightarrow &42& \longleftrightarrow &84&&176& \longleftrightarrow &352\\ && & & \updownarrow&&&&&&\updownarrow \\ 3 & \longleftrightarrow & 6 & & 13 & \longleftrightarrow & 26 & & 53 & \longleftrightarrow & 106& & 213 & \longleftrightarrow & 426 \\ & & \updownarrow & & & & \updownarrow & & & & \updownarrow & & & & \updownarrow \\ 2 & \longleftrightarrow & 4 &\longleftrightarrow& 8 & \longleftrightarrow & 16 & \longleftrightarrow & 32 & \longleftrightarrow & 64 & \longleftrightarrow & 128 & \longleftrightarrow & 256 & \longleftrightarrow & 512 \\ & & & & & & \updownarrow & & & & & & & & \updownarrow \\ & & & & 5 & \longleftrightarrow & 10 & \longleftrightarrow & 20 & \longleftrightarrow & 40 & & 77 & \longleftrightarrow & 154 & \longleftrightarrow & 308 \\ \end{array} $$
Your starting-position is a sexual one for two, and you finish where Bradbury's famous work catches fire.
Further (arithmetical) clues to those positions:
Their product is 31119 and their sum is 520.
Bonus: Why $560$, not $600$, say?