If $(x,y,z)$ have a common factor $h$, divide through by this common factor, and so we can assume that $\gcd(x,y,z)=1$. Also wlog assume $x<y<z$. $x+z=2y$, as $x,y,z$ are in arithmetic progression. Hence $\gcd(x,y)=\gcd(z,y)=1$ and $\gcd(x,z) \leq 2$. (*) So $y=2^a$ or $y=3^b$, as $z>1$ and hence it has a prime factor. Case 1: $x=3^a, y=2^b, z=3^c$. By (*), $a=0$ and $x=1$. So $1+3^c=2^{b+1}$. By Mihalescu's theorem, the only solution is when $b=c=1$. This gives us the solution triple $1,2,3$. Case 2: $x=2^a, y=3^b, z=2^c$. By (*), $a \leq 1$. If $a=0$ then $x$ is odd, and using $x+z=2y$, $z$ must be odd. But this is impossible for $z=2^c>x$. Hence $a=1$. Then $x=2$ and $2+2^c=2\cdot3^b$. Dividing, $1+2^{c-1}=3^b$. By Mihalescu's theorem, the solutions are $b=1$, $c=2$ or $b=2$, $c=4$. Hence the three solution triples are $(1,2,3)$, $(2,3,4)$ or $(2,9,16)$. We may multiply back through by $h$, giving the three numbers a common factor. Hence the solutions are $(h,2h,3h)$, $(2h,3h,4h)$ or $(2h,9h,16h)$, where $h=2^m3^n$.

tim1234133 wrote: By Mihalescu's theorem Mihăilescu's theorem is a very deep theorem (only proved in 2002), I cannot believe this is needed to solve the problem. There must be another way to prove this, seeing it's only a British MO problem, no?

You don't need Mihalescu, although it avoids a couple of lines of easy mod-bashing. Case 1: If b>1, the RHS is a multiple of 8, but the RHS is congruent 2 or 4 mod 8, so we cannot have a solution. Case 2: We have $1+2^{c-1}=3^b$ If $c\geq 3$, $b$ must be even considering mod 4, so let $b=2d$. Then $2^{c-1}=(3^d+1)(3^d-1)$ whence it is obvious that d=1 is the only solution, because only 4 and 2 are powers of two differing by 2.