Consider the list $L$ of numbers from which the first player loses; it is easy to see the list starts as $0, 2, 5, 7, \ldots$. The opposite list $W$ of numbers from which the first player wins is the complement of $L$, and for each number $w$ in $W$ there must exist a number $\ell$ in $L$ such that $w-\ell$ is a perfect square $k^2$. Assume $L$ is finite, and $n$ is its largest element. Then the number $n^2+n+1$ must belong to $L$, since $n^2+n+1 < (n+1)^2$, while $n^2+n+1 - k^2 > n$ for $k\leq n$.