This is a follow-up to this MSE question:
How often does $D(n^2) = m^2$ happen, where $D(x)$ is the deficiency of $x$?
Here is my question in this post:
Do there exist examples for which $D(n^2) = m^2$ and $m \mid n$, where $m, n \in \mathbb{N}$ and $m > 1$?
MY ATTEMPT
In the linked MSE question, the following examples for $D(n^2) = m^2$ (without the divisibility constraint) were given:
$$(m_1, n_1) = (19, 46), m_1 \nmid n_1$$ $$(m_2, n_2) = (53, 284), m_2 \nmid n_2$$ $$(m_3, n_3) = (3509, 149728), m_3 \nmid n_3$$ $$(m_4, n_4) = (6479, 242656), m_4 \nmid n_4$$ $$(m_5, n_5) = (1583, 1633), m_5 \nmid n_5$$ $$(m_6, n_6) = (223309, 260495), m_6 \nmid n_6$$
Rephrasing my question a bit, I have:
Do there exist deficient-perfect square numbers whose deficiency is also a square (and not one)?
Added on July 09 2017 I have the following bounds for $D(n^2)/n^2$ in terms of an auxiliary variable $q$:
$$\frac{2}{q+1} \leq \frac{D(n^2)}{n^2} < \frac{2}{q}.$$
The conditions specified in my problem imply that
$$\frac{2}{q+1} = \frac{m^2}{n^2}.$$
Added on July 12 2017 - MOTIVATION
When $N=q^k n^2$ is an odd perfect number, then the following equations hold: $$\gcd\left(n^2,\sigma(n^2)\right)=\frac{D(n^2)}{\sigma(q^{k-1})}=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$ (This is proved in this preprint, to appear in NNTDM.)
In particular, when the Descartes-Frenicle-Sorli conjecture that $k=\nu_{q}(N)=1$ is true, then $D(n^2) \mid n^2$.
In the other direction, Broughan, et. al (page 7, Lemma 8) proved that if $$\frac{\sigma(N/q^k)}{q^k}=\frac{\sigma(n^2)}{q^k}$$ is a square, then $k=1$.
Putting these all together, under the assumption that $\sigma(n^2)/q^k$ is a square, we have that $$1 < m^2 = D(n^2) \mid n^2.$$
Now, MSE user Matthew Conroy checked all $n < {10}^7$ and found only one ODD deficient-perfect square (other than $1$), namely $n^2 = {3003}^2 = 9018009$, with deficiency $819$. (Notice that the deficiency is not a square.)
All other deficient-perfect squares are powers of $2$, up to $n < {10}^7$. Indeed, if $n = 2^k$ for some integer $k \geq 0$, then $n^2$ is deficient-perfect with deficiency $1$.