a) Let's go for an induction procedure : First note that 1 and 4 are the first two terms of the arithmetic progression $1+3n$ where $n=0,1,2,...$. Now suppose that for some $k$, you have an arithmetic progression $x+nr$ and that for $n=0,1,...,k-1$ we have $x+nr = a_n^{b_n}$ for some positive integers $a_n,b_n$ and $b_n >1$. Let $q= lcm \{b_1,...,b_n \}$. Let $t$ be any positive integer. Then, using that $t^q(x+nr) = t^qx + n(t^qr)$, we see that $ \{t^q(x+nr) \}$ is an arithmetic progression. Moreover for $n=0,1,...,k-1,$ we have let $q=b_nc_n$, we then have $t^q(x+nr) = t^{b_nc_n}a_n^{b_n} = (t^{c_n}a_n)^{b_n}$ is a perfect power. Now choose $t=x+kr$, clearly $t^q(x+kr)$ is a perfect power too, and so we have an arithmetic progression formed with perfect power of kength $k+1$. It follows that we may find arithmetic progressions of perfect powers with arbitrary long, but finite, length. b) Suppose that $x+nr$ is an infinite arithmetic progresion form with perfect powers. Let $d = gcd(x,r)$, $x=dy$ and $r=ds$, where $y,s$ are coprimes. Then for each $n$, we have $x+nr = d(y+ns)$. Now, from the well-known theorem of Dirichlet about primes contained in the arithmetic progressions, we know that there are an infinite number of primes of the form $y+ns$. Thus, we may find $n$ such that $y+ns$ is a prime $p$ which does not divide $d$. In that case $d(y+ns)$ is a term of the sequence, it is divisible by $p$ but not by $p^2$. Therefore it cannot be a perfect power. A contradiction. Pierre.

It was in our exams last year (Iran 2003)

for a) you can find number $d$ such that all numbers $d, 2d... nd$ are all perfect powers. Solution is boring, based on CRT. First you have to choose all primes less or equal than $n$ and carefully choose their exponents in factorization of $d$. Choosing is similar like http://www.artofproblemsolving.com/Forum/viewtopic.php?p=117503&sid=c792bd67b26d3da8c862eb112a28e1fb#p117503 - here of course you need to use $n^k$ primes- there only $ n^2$ were enough.