Suppose the sequence does contain a square.
Let say x_{k} be the first square, and let it equal r_{2}.
We have r^{2} = x_{k} = ak + b.

Now, consider that:

(r + ta)^{2} = r^{2} + 2rta + t^{2}a^{2} = ak + b + 2rta + t^{2}a^{2} = a(k + 2rt + t^{2}a) + b.

So (r + ta)^{2} = x_{k + 2rt + t2} is also in the sequence for any positive integer t, and the sequence therefore contains infinitely many squares.