We can easily extend the definition of a square polynomial to polynomials with complex coefficients. In all the arguments below we consider polynomials with complex coefficients.

Lemma: If $p(x)$ is a square and $\alpha$ is a non-zero complex number then $p(x) - \alpha$ is not a square.

Proof of Lemma: Suppose $p(x) = q(x)^2$ and $p(x) - \alpha = r(x)^2$, with both $q(x)$ and $r(x)$ being non-constant polynomials. Then $\alpha = (q(x) - r(x))(q(x) + r(x))$. Clearly, either $q(x) - r(x)$ or $q(x) + r(x)$ is not a constant polynomial, and hence a contradiction. This proves the lemma.

Continuation of the solution: We can write $f(x)$ as $f_1(x)^2(x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_k)$, where $f_1(x)$ is a polynomial and $\alpha_1, \alpha_2, \cdots, \alpha_k$ are distinct complex numbers. Then $f(g(\tau)) = f_1(g(\tau))^2(g(\tau) - \alpha_1) \cdots (g(\tau) - \alpha_k)$ is a square. It follows that $(g(x) - \alpha_1)(g(x) - \alpha_2) \cdots (g(x) - \alpha_k) = h(x)^2$ for some polynomial $h(x)$. Let $\beta$ be such that $g(\beta) = \alpha_1$. Then $h(\beta) = 0$ and hence $h(x) = (x - \beta)h_1(x)$. Note that $g(\beta) - \alpha_i = 0$ for any $i = 1$. Therefore it follows that $(x - \beta)^2$ divides $g(x) - \alpha_1$. We can then conclude that $g(x) - \alpha_1$ is a square. Similarly, $g(x) - \alpha_i$ is a square for $i = 1, 2, \cdots, k$. By the above lemma, it follows that $k = 1$, so $f(x) = f_1(x)^2(x - \alpha)$ and $g(x) = g_1(x)^2 + \alpha$ for some non-zero complex number $\alpha$. Therefore $g(f(x)) = g_1(f(x))^2 + \alpha$ and hence by the above lemma it follows that $g(f(x))$ is not a square. This completes the proof.