Assume $a_{n+1}$ is a square. Then considering $a_n^5+487=g^2$ ($a_{n+1}=g^2$) modulo 8 we get $g\equiv 0( \mod 2)$. Then we find $a_{n+2}\equiv 7 (\mod 8)$, $a_{n+3}\equiv 6 (\mod 8)$, $a_{n+4}\equiv 7 (\mod 8)$. So there is no perfect square after $a_{n+1}$. Hence there can't be 3 (or more) perfect squares in this sequence. Assume there are two prefect squares. From the above it follows that $a_0$ is a perfect square. $m=h^2$. Assume $a_1$ is not a perfect square. If $a_0$ is even, then we get $a_{1}\equiv 7 (\mod 8)$, $a_{2}\equiv 6 (\mod 8)$, $a_{3}\equiv 7 (\mod 8)$ and there are no other perfect squares. If $a_0$ is odd, then it is 1 modulo 4, and it follows $a_1\equiv 0 (\mod 8)$, $a_{2}\equiv 7 (\mod 8)$, $a_{3}\equiv 6 (\mod 8)$, $a_{4}\equiv 7 (\mod 8)$, and since $a_1$ is not a perfect square, there are no other perfect squares. Thus it remains $a_1$ is a perfect square. $a_1=r^2$. Now we have $r^2=h^{10}+487$, $(r+h^5)(r-h^5)=487$. Since 487 is prime we get $r=244$, $h^5=243$. Finally $m=9$.

Clearly every two consecutive terms have different parity. Observe that if $a_k$ is even for some $k\geq 0$, then $a_n$ is not a perfect square for all $n\geq k$. Indeed, we must have $a_{k+1} \equiv 3 \pmod 4$, $a_{k+2} \equiv 2 \pmod 4$ and in general $a_{k + 2t-1} \equiv 3, a_{k+2t} \equiv 2 \pmod 4$ for all $t\geq 1$, while a perfect square can give remainder $0$ or $1$ mod $4$ only. Therefore, as exactly one of $a_0$ and $a_1$ is even, we obtained that $a_n$ cannot be a square for all $n\geq 2$. In particular, the sequence cannot contain more than two perfect squares - and if there are two, they must be $a_0$ and $a_1$. If $a_0 = m = x^2$ and $a_1 = y^2$ for $x,y > 0$, then $y^2 = x^{10} + 487$, equivalent to $(y - x^5)(y + x^5) = 487$. Since $487$ is prime, we can only have $y - x^5 = 1$, $y + x^5 = 487$, thus $x^5 = 243$, $x=3$ and $a_0 = m = 9$ is the only solution.