If there is a perfect square in the sequence, we may assume that it's $a_{1}$. Assume for the sake of contradiction that $a_{n}$ is a perfect square for some $n\geq 3$. So, $a_{n}=x^2=a_{n-1}^3 + 103$. If $a_{n-1}$ is even, $x^2\equiv 7 (mod 8)$, contradiction. If it is odd, then $a_{n-2}^3 = a_{n-1} - 103$ is even, so $a_{n-1}\equiv 7 (mod 8)$ and $x^2\equiv 6 (mod 8)$, contradiction. It remains to see if $a_1$ and $a_2$ can be both perfect squares. Let $a_1 = m^2$ and $a_2 = k^2$. Then $k^2 = m^6 + 103$, so $(k-m^3)(k+m^3)=103$. Since 103 is prime, $k-m^3=1$, so $1+2m^3=103$, which does not lead to a solution.

This problem is pretty funny ... it's easy to check that no two consecutive terms can be perfect squares. Now, note that squares are $0$, $1$, or $4$ mod $8$. After a perfect square appears, the next term must be $-1$ or $0$ mod $8$, and thereafter all terms are $-1$, $-2$ modulo $8$, so no more squares.

Edit: Fakesolve lol We claim that for $k \geq 2$, $a_k$ can't be a square. Taking mod 7, $y^2 \equiv x^3 + 103$(mod 7) isn't possible. This is because $y^2$(mod 7) can only be 0,1,2,4 and $x^3$(mod 7 ) can only be -1,0,1. Thus, since only $a_1$ can be a square, there is at most one square.

AwesomeYRY wrote: We claim that for $k \geq 2$, $a_k$ can't be a square. Taking mod 7, $y^2 \equiv x^3 + 103$(mod 7) isn't possible. This is because $y^2$(mod 7) can only be 0,1,2,4 and $x^3$(mod 7 ) can only be -1,0,1. Thus, since only $a_1$ can be a square, there is at most one square. $(x, y)=(-1,2)$ satisfy $y^2\equiv x^3+103 $(mod 7). So maybe mod 7 cannot finish the problem I think.