Since every move decreases the number of heaps by $2$, we must start with an odd number $k=2n+1$ of heaps. Then the game ends after $n$ moves, which altogether remove $1+2+\cdots+n$ stones from the total number $1+2+\cdots+(2n)+(2n+1)$. Hence \[p=\frac12(2n+1)(2n+2)-\frac12n(n+1)=\frac12(n+1)(3n+2)=\frac{1}{16}(2k+2)(3k+1).\] Now the greatest common divisor of $2k+2$ and $3k+1$ divides $3(2k+2)-2(3k+1)=4$. Since the product of these two terms should be a square, this means that they either (1) both are squares, or (2) otherwise $2k+2=2x^2$ and $3k+1=2y^2$. Now (2) is impossible, since the square $y^2$ is either $0$ or $1$ modulo $3$, so that $2y^2$ is $0$ or $2$ modulo $3$. Hence case (1) must hold true, and that's exactly what we wanted to prove. To find the smallest $k\ge3$ for which $p$ is a square, we must find the smallest solution of $2k+2=x^2$ and $3k+1=y^2$. We deduce $3x^2-2y^2=4$. Then $x=2a$ and $y=2b$ must be even, and we get $3a^2-2b^2=1$. Considering this equation modulo $8$ shows that $a$ and $b$ both must be odd. Now $b=1$ yields the infeasible $k=1$, and $b=3,5,7,9$ yield non integer values for $a$. The smallest solution has $b=11$ and $a=9$, which yields the answer $k=161$.