= Solution
Let $M=|m|$ for integer $m$. In the even or odd <power series> whose parity matches $M$, the numerator in the <recurrence relation> vanishes at the coefficient of $x^M$. All coefficients of lower powers in that parity are nonzero, and all higher ones vanish. Therefore a <polynomial> solution exists, of <polynomial degree> $M$, containing
$$
\boxed{\left\lfloor\frac M2\right\rfloor+1\text{ nonzero terms}.}
$$
For nonnegative $m$ these are the <Chebyshev polynomials> $T_m$. The normalized low-degree cases follow directly from the <recurrence relation>:
$$
\begin{aligned}
m=0:&\quad y=a_0,\\
m=1:&\quad y=a_1x,\\
m=2:&\quad y=a_0(1-2x^2),\\
m=3:&\quad y=a_1(x-\tfrac43x^3).
\end{aligned}
$$
Enforcing $T_m(1)=1$ gives
$$
\boxed{T_0=1,\qquad T_1=x,\qquad T_2=2x^2-1,\qquad T_3=4x^3-3x.}
$$
To see that normalization at one is always possible, set $x=\cos\theta$ in the <Chebyshev differential equation>. For $Y(\theta)=y(\cos\theta)$, the differential expression $(1-x^2)y''-xy'$ becomes $Y''$. Thus $Y=\cos(M\theta)$ gives $T_M(x)=\cos(M\arccos x)$ on $[-1,1]$ and has value one at $x=1$. It is a <polynomial> because the <cosine addition formula> gives $T_{M+1}=2xT_M-T_{M-1}$. Negative integer $m$ produces the same solution as $M=|m|$.
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