First, we prove a lemma.

Lemma The following Diophantine equation
$$x^2 + 11y^2 = 4m$$
has a solution $(x_0, y_0)$, such that either $x_0$, $y_0$ are odd numbers or $x_0$, $y_0$ are even numbers with $x_0 \equiv (2k+1)y_0 \pmod m$.

Consider the expression $x + (2k+1)y$, where $x$, $y$ are integers, and $0 \le x \le 2\sqrt{m}$, $0 \le y \le \frac{\sqrt{m}}{2}$.

There are $([2\sqrt{m}]+1)\left\lfloor\frac{\sqrt{m}}{2}\right\rfloor+1 \ge m$ such expressions. So there exist integers $x_1$, $x_2 \in [0, 2\sqrt{m}]$, $y_1$, $y_2 \in [0, \frac{\sqrt{m}}{2}]$, such that $(x_1, y_1) \ne (x_2, y_2)$, and
$$x_1 + (2k+1)y_1 \equiv x_2 + (2k+1)y_2 \pmod m,$$
So
$$x \equiv (2k+1)y \pmod m, \quad \text{where} \quad x = x_1 - x_2, \ y = y_2 - y_1.$$
This means
$$x^2 \equiv (2k+1)^2 y^2 \equiv -11y^2 \pmod m,$$
that is, $x^2 + 11y^2 = tm$ for some integer $t$.
Since $|x| \le 2\sqrt{m}$, $|y| \le \frac{\sqrt{m}}{2}$, we have
$$x^2 + 11y^2 < 4m + \frac{11}{4}m < 7m.$$
So $1 \le t \le 6$.
As $m$ is an odd number, obviously the equations $x^2 + 11y^2 = 2m$, and $x^2 + 11y^2 = 6m$ have no integer solution.

(1) If $x^2 + 11y^2 = m$, then $x_0 = 2x$, $y_0 = 2y$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.

(2) If $x^2 + 11y^2 = 4m$, then $x_0 = x$, $y_0 = y$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.

(3) If $x^2 + 11y^2 = 3m$, then $(x \pm 11y)^2 + 11(x \mp y)^2 = 9 \cdot 4m$.
First, assume that $3 \mid m$. If $x \not\equiv 0 \pmod{3}$, $y \not\equiv 0 \pmod{3}$, and $x \not\equiv y \pmod{3}$, then
$$x_0 = \frac{x - 11y}{3}, \quad y_0 = \frac{x + y}{3}$$
is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.
If $x \equiv y \not\equiv 0 \pmod{3}$, then
$$x_0 = \frac{x + 11y}{3}, \quad y_0 = \frac{y - x}{3}$$
is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.
Now suppose $3 \mid m$. Then the above are still integer solutions. If $x^2 + 11y^2 = 4m$ has an even integer solution $x_0 = 2x_1$, $y_0 = 2y_1$, then
$$x_1^2 + 11y_1^2 = m \Leftrightarrow 36m = (5x_1 \pm 11y_1)^2 + 11(5y_1 \mp x_1)^2.$$
Since one of $x_1$, $y_1$ is even and the other is odd, so $5x_1 \pm 11y_1$, $5y_1 \mp x_1$ are odd numbers.
If $x_1 \equiv y_1 \pmod{3}$, then $x_0 = \frac{5x_1 - 11y_1}{3}$, $y_0 = \frac{5y_1 + x_1}{3}$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.
If $x_1 \not\equiv y_1 \pmod{3}$, then $x_0 = \frac{5x_1 + 11y_1}{3}$, $y_0 = \frac{5y_1 - x_1}{3}$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.

(4) If $x^2 + 11y^2 = 5m$, then $25 \cdot 4m = (3x \mp 11y)^2 + 11(3y \pm x)^2$.
When $5 \nmid m$, if
$$x \equiv \pm 1 \pmod{5}, \quad y \equiv \mp 2 \pmod{5}, \quad \text{or } x \equiv \pm 2 \pmod{5}, \quad y \equiv \pm 1 \pmod{5},$$
then
$$x_0 = \frac{3x - 11y}{5}, \quad y_0 = \frac{3y + x}{5}$$
is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.
If $x \equiv \pm 1 \pmod{5}$, $y \equiv \pm 2 \pmod{5}$, or $x \equiv \pm 2 \pmod{5}$, $y \equiv \mp 1 \pmod{5}$, then
$$x_0 = \frac{3x + 11y}{5}, \quad y_0 = \frac{3y - x}{5}$$
is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.

When $5 \mid m$, then the above are still integer solutions. If $x^2 + 11y^2 = 4m$ has an even integer solution $x_0 = 2x_1$, $y_0 = 2y_1$, then
$$x_1^2 + 11y_1^2 = m, \quad x_1 \not\equiv y_1 \pmod 2,$$
and we have $100m = (x_1 \mp 33y_1)^2 + 11(y_1 \pm 3x_1)^2$.
If $x_1 \equiv y_1 \equiv 0 \pmod{5}$, or $x_1 \equiv \pm 1 \pmod{5}$, $y_1 \equiv \pm 2 \pmod{5}$, or
$$x_1 \equiv \pm 2 \pmod{5}, \quad y_1 \equiv \mp 1 \pmod{5},$$
then $x_0 = \frac{x_1 - 33y_1}{5}$, $y_0 = \frac{y_1 + 3x_1}{5}$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.
If $x_1 \equiv \pm 1 \pmod{5}$, $y_1 \equiv \mp 2 \pmod{5}$, or $x_1 \equiv \pm 2 \pmod{5}$, $y_1 \equiv \pm 1 \pmod{5}$, then
$$x_0 = \frac{x_1 + 33y_1}{5}, \quad y_0 = \frac{y_1 - 33x_1}{5}$$
is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.

The lemma is proved.

From the lemma, if $x^2 + 11y^2 = 4m$ has a solution $(x, y)$ with $x$, $y$ being odd numbers, then it has a solution $(x_0, y_0)$ with $x_0$, $y_0$ being even numbers satisfying $x_0 \equiv (2k+1)y_0 \pmod m$.

Let $l = 2k + 1$, the solution of the quadratic equation
$$mx^2 + ly_0x + ny_0^2 - 1 = 0$$
is
$$x = \frac{-ly_0 \pm \sqrt{l^2 y_0^2 - 4mny_0^2 + 4m}}{2m} = \frac{-ly_0 \pm x_0}{2m}.$$
So the equation
$$mx_1^2 + ly_0x_1 + ny_0^2 - 1 = 0$$
has at least an integer solution $x_1$, i.e.
$$mx_1^2 + ly_0x_1 + ny_0^2 - 1 = 0.$$
This indicates that $x_1$ is an odd number. Now, from this equation, it follows that
$$(2ny_0 + lx_1)^2 + 11x_1^2 = 4n.$$
This means $x^2 + 11y^2 = 4n$ has a solution $(x, y)$ with $x$, $y$ being odd numbers, where $x = 2ny_0 + lx_1$, $y = x_1$.

Therefore, at least one of the equations $x^2 + 11y^2 = 4m$ or $x^2 + 11y^2 = 4n$ has a solution $(x, y)$ with $x$, $y$ odd.