Let $C = \{ z \in \mathbb{C} : |z| = 1 \}$ and $D = \{ z \in \mathbb{C} : |z| < 1 \}$ be the unit circle and open unit disk. We will assume $a \ne \pm 1$ as their cases are trivial.
Your statement is true. We are going to prove following generalization:
For $\alpha_1, \alpha_2, \ldots, \alpha_m \in D$, define
$f(z) = \prod\limits_{k=1}^m(z - \alpha_k)$ and $g(z) = \prod\limits_{k=1}^m (1-\bar{\alpha}_k z)$.
The polynomial $f(z) - g(z)$ has all its roots belong to $C$.
Consider their ratio $h(z) = \frac{f(z)}{g(z)}$.
Since all $|\alpha_k| < 1$, $g(z)$ is never zero over $C$ and $h(z)$ is well defined there.
For $z \in C$, we have
$$|h(z)|
= \prod\limits_{k=1}^m \left|\frac{z-\alpha_k}{1-\bar{\alpha}_k z}\right|
= \prod\limits_{k=1}^m \left|\frac{z-\alpha_k}{(\bar{z} - \bar{\alpha}_k)
z}\right|
= 1$$
The ratio $h(z)$ maps $C$ to $C$.
For each factor $\frac{z-\alpha_k}{1-\bar{\alpha}_k z}$, when $z$ move long $C$ once, the factor move along $C$ also once. This implies their product $h(z)$ move along $C$ exactly $m$ times. As a result, we can find $m$ distinct $\theta_1, \ldots, \theta_m \in [ 0, 2\pi )$ such that
$$h(e^{i\theta}) = 1 \iff f(e^{i\theta}) - g(e^{i\theta}) = 0$$
Polynomial $f(z) - g(z)$ has at least $m$ distinct roots over $C$. Since degree of $f(z) - g(z)$ is $m$, counting multiplicity, it has exactly $m$ roots in $\mathbb{C}$. This means above $m$ roots on $C$ is all the roots of $f(z) - g(z)$ and all of them are simple.
On the special case $m = n + 1$ and $(\alpha_1,\alpha_2,\ldots,\alpha_m) = (a,0,\ldots,0)$ where $a \in (-1,1)$.
Polynomial $f(z) - g(z)$ reduces to
$$z^n(z - a) - (1-az) = z^{n+1} - a z^n + az - 1 = P(z)$$
and your statement follows.
IMHO, this is a good chance to introduce the concept of winding number to the students. If they are not ready for that. A standalone proof for the original statement (again $a \ne \pm 1$) goes like this.
When $a \in (-1,1)$, parameterize $C$ by $[0,2\pi) \ni \theta \mapsto z \in C$. We have
$$P(z) = z^{n+1} - az^n + az - 1 = 2ie^{i\frac{(n+1)\theta}{2}}
\left[\sin\frac{(n+1)\theta}{2} - a\sin\frac{(n-1)\theta}{2}\right]$$
Let's call what's inside the square bracket as $I(\theta)$.
When $a$ is real, $I(\theta)$ is clearly real and $\theta = 0$ is a root of it.
Let $\theta_k = \frac{(2k+1)\pi}{n+1}$ for $k = 0,\ldots,n$. When $a \in (-1,1)$, it is easy to see $I(\theta_k)$ is positive for even $k$ and
negative for odd $k$. This means $I(\theta)$ has $n$ more roots. One root
from each interval $(\theta_{k-1},\theta_k)$ for $k = 1,\ldots, n $.
As a result, $I(\theta)$ has at least $n+1$ roots over $[0,2\pi)$. This is equivalent to $P(z)$ has at least $n+1$ roots on $C$. Once again, since $P(z)$ has degree $n+1$, these are all the roots it has.
Let $t^4 - 2xt^2 + y^2 = 0$ be our equation, with roots $x_1, x_2, x_3, x_4 \in \mathbb{R}$ such that they are in arithmetic progression. Because of this, then there exists $\alpha, r \in \mathbb{R}$ such that $x_1 = \alpha - 3r, x_2 = \alpha -r, x_3 = \alpha + r, x_4 = \alpha + 3r$, and then our solutions are in an arithmetic progression of ratio $2r$.
Notice that $S_1 =x_1 + x_2 + x_3 + x_4 = 4\alpha$ by our assumption. By using the Viète sum, we obtain that $S_1 = 0$, therefore $\alpha = 0$, and our solutions become $x_1 = -3r, x_2 = -r, x_3 = r, x_4 = 3r$.
The Viète product of all solutions is $y^2$, therefore $9r^4 = y^2 \Leftrightarrow |y| = 3r^2$. If $r = 0$, then $x = y = 0$, and the solutions are in a constant arithmetic progression, $x_1 = x_2 = x_3 = x_4 = 0$. Suppose $r \neq 0$. Therefore, because $r$ is a solution, plugging it into our original equation, we obtain $r^4 - 2xr^2 + 9r^4 = 0 \Leftrightarrow x = 5r^2$. Finally, we obtain the following solutions:
$(x, y) \in \{(5r^2, 3r^2), (5r^2, -3r^2) : r \in \mathbb{R}\} = \{(5\beta, 3\beta), (5\beta, -3\beta) : \beta \geq 0\}$
Note: Whenever you have a polynomial with solutions either in an arithmetic or geometric progression, the idea is to consider the solutions in such a way so that when adding them (A.P.) or multiplying them (G.P.), you are only obtaining an equality in one term.
Best Answer
I assume that by "roots in arithmetic progression" you mean that the four roots are of the form $$ \alpha,\quad\alpha+k,\quad\alpha+2k,\quad\alpha+3k $$ for some $\alpha$ and some $k\neq0$.
Now we know that $0$ and $m$ are roots. Since the coefficient of $x^3$ is $-m$ the sum of the roots must be $m$ which gives $\alpha=-m/2$ and $k=m/2$, i.e the four roots are $$ -\frac m2,\quad0,\quad\frac m2,\quad m. $$ Since one of the roots is $0$ the coefficient of $x$ is the negative of the product of the three non-zero roots, i.e. $$ 2m^2=\frac{m^3}4 $$ whose only non-zero solution is $m=8$.
On the other hand, if you do not assume that the four roots are distinct, since anyway the four roots are $\{-\beta,0,\beta,m\}$ for some $\beta$ one gets two extra-cases, namely $\beta=0$ and $\beta=m$.
Working again with the linear coefficient we soon get that