Because $b|a+1$, $c|b+1$ and $a|c+1$, it follows $a\leq c+1\leq b+2\leq a+3$. Now it's just some casework. Wlog either $a\leq b\leq c$ or $a\leq c\leq b$. I)we must have $b\leq c\leq b+1$. The case where $b=c$ only works if $b=c$ and implies $a=1$ or $a=2$ (the latter doesn't work). The case where $c=b+1$ implies $b-1\leq a\leq b$; if $a=b-1$ then $b=4$ and $a=3$ and $c=5$; the case where $a=b=1$ and $c=2$ II) we have $a\leq c\leq b\leq a+1$. If $b=a+1$ then $a\leq c+1\leq a+3$ which implies $a\leq c\leq a+1$. If $c=a$ then $a=c=1$ and $b=2$. If $a=b=c$ then the only one is $a=b=c=1$. Thus the only solutions are $(2,1,1)$, $(1,1,1)$,$(3,4,5)$ and cyclic permutations.