Let $f(x,y)$ be a binary quadratic form with co-prime integer coefficients. We say that $f$ is a *proper subform* of $g(x,y)$ if there exists an integer matrix $A = \left(\begin{smallmatrix} a_1 & a_2 \\ a_3 & a_4 \end{smallmatrix}\right)$ with $|\det A| > 1$ such that

$$\displaystyle f(x,y) = g(a_1 x + a_2 y, a_3 x + a_4 y).$$

For example, the form $f(x,y) = 4x^2 + 4xy + 5y^2$ is a proper subform of $g(x,y) = x^2 + y^2$, since

$$\displaystyle 4x^2 + 4xy + 5y^2 = (2x + y)^2 + (2y)^2.$$

If $f$ is a proper subform, then the discriminant of $f$ is divisible by $\det(A)^2$, so it is not square-free. My question is the converse: suppose that $\Delta(f)$ is divisible by an odd square $m^2$. Is $f$ a proper subform of another form $g$?