Does perfect square imply perfect square polynomial?

By completing the square, we simply need to consider polynomials of the form $x^2 + \frac{n}{m}$ , where $m,n$ are coprime integers.

With $x = 0$ , we get that $\frac{n}{m}$ is a square of a rational number, and hence $nm$ is a perfect square.
With $x = \frac{1}{m}$ , we get that $\frac{nm + 1 } {m^2}$ is a square of a rational number, and hence $nm +1$ is a perfect square.

Since the only 2 perfect squares that differ by one are 0 and 1, hence we conclude that $nm = 0$ . Since $m \neq 0$ , thus $n = 0$ , and hence we conclude that $f(x) = x^2$ which is a square of a polynomial.