The nilpotency class of $G_d$ is indeed always 3. One way to see this is to rewrite the presentation of $G_d$ in such a way to exhibit that it is a polycyclic group. For this purpose, let $z:=[x,y]$ and $w:=y^{2^{d-1}}=x^{2^d}$. Clearly $w$ lies in the center of $G_d$. With a little more effort we see that $$ G_d \cong \langle x, y, z, w\mid x^4 = y^2, y^{2^{d-1}} = w, z^2 = w^2 = 1 ; y^x = yz, z^x = zw, z^y = w^x=w^y=w^z=1 \rangle $$

This is indeed a polycyclic presentation, with relative order $4, 2^{d-1}, 2, 2$. (Thus the group has order $4* 2^{d-1}* 2* 2=2^{d+3}$).

But now it is easy to read off that $[G_d,G_d] = \langle z, w\rangle$, and thus $[[G_d,G_d],G_d]=\langle w \rangle$, which is central. Hence the nilpotency class is 3.

The nilpotency class of $G_d$ is indeed always 3. One way to see this is to rewrite the presentation of $G_d$ in such a way to exhibit that it is a polycyclic group. For this purpose, let $z:=[x,y]$ and $w:=y^{2^{d-1}}=x^{2^d}$. Clearly $w$ lies in the center of $G_d$. With a little more effort we see that $$ G_d \cong \langle x, y, z, w\mid x^4 = y^2, y^{2^{d-1}} = w, z^2 = w^2 = 1 ; y^x = yz, z^x = zw, z^y = z, w^x=w^y=w^z=w \rangle $$

This is indeed a polycyclic presentation, with relative order $4, 2^{d-1}, 2, 2$. (Thus the group has order $4* 2^{d-1}* 2* 2=2^{d+3}$).

But now it is easy to read off that $[G_d,G_d] = \langle z, w\rangle$, and thus $[[G_d,G_d],G_d]=\langle w \rangle$, which is central. Hence the nilpotency class is 3.