By induction we can show that $a_n > 0$ for all positive integer $n$. Suppose that there exists a positive integer $n$ such that $a_n$ is a perfect square.
$$a_n = a_{n-1}^2 + 15a_{n-1} = a_{n-1}(a_{n-1} + 15)$$
so $a_{n-1} \mid a_n$ for all positive integers $n$. Thus, it follows that
$$a_1 = 45 \mid a_n, \quad \text{for all positive integers } n.$$
Let $k > 1$ be the smallest positive integer such that $a_k$ is a perfect square and $a_{k-1} = 45x$ with $x$ a positive integer. We have
$$a_k = 45x(45x + 15) = 15^2 \cdot 3x(3x + 1)$$
is a perfect square. On the other hand, $(3x, 3x + 1) = 1$ implies $3x$ and $3x + 1$ are perfect squares, so $3x = 0$ and $3x + 1 = 1$. It is impossible since $a_n > 0$ for all $n$.
$\square$