Pretty sure there is a shorter solution, but this should work.

Assume for the sake that it is possible. Denote the number of 5 petals by $x$, 4 petals by $y$. $$x+y=m^2, \ 5x+4y=n^2$$By our assumption, $x \mid y \implies y = xk$ $x = n^2 - 4m^2$ and $y = 5m^2 - n^2$. Hence we have $5m^2 - n^2 = k(n^2 - 4m^2)$. This can be rewritten as $\frac{4k + 5}{k + 1} = \frac{n^2}{m^2}$. Now notice that, since $k$ is integer, we have $\gcd(4k + 5, k + 1) = \gcd(1, k + 1) = 1$, hence both $4k + 5$ and $k + 1$ are actually square of integers. This leads to $4k + 5 = u^2$ and $k + 1 = v^2$, or $(2v)^2 + 1 = u^2$. This is impossible, since two positive squares cannot differ by $1$.

Wait... but if the number of 5-flowers is a perfect square, and if the number of 4-flowers is 0, then all the conditions are satisfied

Não amigo. Neste caso o número total de pétalas seria 5k^2 que não é um quadrado perfeito.