The point $P = (-1,0)$ lies on the ellipse $4x^2 + y^2 = 4.$  Let $Q$ be another point on this ellipse, and let $d$ be the maximum length of $\overline{PQ}.$  Find $d^2.$
Let $Q = (x,y).$  From the given information, $y^2 = 4 - 4x^2.$  Therefore,
\begin{align*}
PQ^2 &= (x + 1)^2 + y^2 \\
&= x^2 + 2x + 1 + 4 - 4x^2 \\
&= -3x^2 + 2x + 5 \\
&= -3 \left( x - \frac{1}{3} \right)^2 + \frac{16}{3}.
\end{align*}This is maximized when $x = \frac{1}{3},$ and $d^2 = \boxed{\frac{16}{3}}.$