= Solution
First classify nonzero <polynomial> solutions of the <differential equation> itself. If $y$ has degree $d$ and leading <coefficient> $a_d\ne0$, the <coefficient> of $x^{d+1}$ is $(d+\lambda)a_d$. Hence $\lambda=-d$ is necessary.
An analytic solution at zero belongs to the exponent-zero Frobenius branch. Its odd-power <coefficients> vanish, because $a_1=0$ and the <recurrence relation> preserves <parity>. For the even-power <coefficients>,
$$
a_{2j+2}=-\frac{2j+\lambda}{(2j+2)(4j+1)}a_{2j}.
$$
All denominators are nonzero for $j\geq0$. With $\lambda=-2m$, the <coefficients> through $a_{2m}$ are nonzero when $a_0\ne0$, the next <coefficient> vanishes, and every subsequent <coefficient> vanishes. This proves sufficiency as well as the restriction to even <polynomial> degree:
$$
\boxed{\lambda=-2m,\qquad m=0,1,2,\ldots,\qquad \deg y=2m=-\lambda.}
$$
For example, setting $a_0=1$ gives $1$ when $\lambda=0$, $1+x^2$ when $\lambda=-2$, and $1+2x^2+x^4/5$ when $\lambda=-4$. A <polynomial> cannot instead begin with a <positive integer> power: its lowest power must have an <indicial exponent>, and the only <integer> <indicial exponent> is zero. This also proves that no odd-degree family was missed. The finite-series mechanism is a <terminating Frobenius series>.
\b[If the additional endpoint conditions from part (b) are imposed on this part too, there is no nonzero <polynomial> solution:] every nonzero <polynomial> just classified has $y(0)=a_0\ne0$. The displayed family and degree answer therefore concern the <differential equation> without those extra endpoint conditions.
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