If there is one perfect square in an arithmetic progression, then there are infinitely many

Note that $$(m+d)^2=m^2+(2m+d)\cdot d$$

Let $n^2$ be the known square.

Then $n^2+kd=m^2$ is equivalent to $kd=(m-n)(m+n)$. You can take $m-n$ to be any multiple of $d$, and $k$ follows.

Starting from where you left, suppose the $k$th term is a perfect square such that $$a_k = a_1 + (k-1)d = p^2$$

Now add $2pmd + m^2d^2 $ to both sides where $m$ is a natural number giving us, $$\Rightarrow a_1 + (k-1)d + 2pmd + m^2d^2 = p^2 + 2pmd + m^2d^2 $$ $$\Rightarrow a_1 + [(k-1)+2pm+m^2]d = (p+md)^2$$

The RHS is a perfect square, and the left side is the $(k-1+2pm+m^2)$th term for infinitely many values of $m $. Hope it helps.