There are no solutions to this equation -- the point is that for $d\ge 5$ the term $1+N(d,q)$ is (much) larger than $q^{d+1}$, while for $d=2$, $3$ or $4$ we get a polynomial equation whose solutions are easily determined. To see this, first note that
$$
\binom{d}{k}_q =\frac{(1-q^d)(1-q^{d-1})\cdots (1-q^{d-k+1})}{(1-q)(1-q^2)\cdots(1-q^k)} = \frac{(q^d-1)}{(q^k-1)}\frac{(q^{d-1}-1)}{(q^{k-1}-1)}\cdots \frac{(q^{d-k+1}-1)}{(q-1)},
$$
and since $(x-1)/(y-1) \ge x/y$ if $x \ge y >1$, it follows that
$$
\binom{d}{k}_q \ge \frac{q^d}{q^k} \frac{q^{d-1}}{q^{k-1}}\cdots \frac{q^{d-k+1}}{q} = q^{k(d-k)}.
$$
Thus if $d\ge 5$, then
$$
N(d,q) \ge \binom{d}{2}_q \ge q^{2(d-2)} \ge q^{d+1},
$$
and there are no solutions to the given equation. (In fact, much more is known about $\binom{d}{k}_q$ -- namely this is a polynomial in $q$ of degree $k(d-k)$ with leading coefficient $1$, and with all other coefficients being non-negative integers. See for example Amritanshu Prasad's answer to the linked MO question Sum of Gaussian binomial coefficients. This gives an alternative proof that $\binom{d}{k}_q \ge q^{k(d-k)}$ for all non-negative $q$.)

This leaves us with the cases $d=2$, $3$ and $4$. When $d=2$, we see that $N(2,q)= 3+q$ and for all prime powers $q$ we see that $1+N(2,q) < q^3$. If $d=3$ then $N(3,q)=2+2(1+q+q^2)$ so that the equation is $5+2(q+q^2)=q^4$ which can readily be checked to have no prime power solutions. Finally if $d=4$ then $N(4,q) = 2+2(1+q+q^2+q^3)+(1+q^2)(1+q+q^2) = q^4+3q^3+4q^2+3q+5$, and the equation is
$$
q^4+3q^3+4q^2+3q+6=q^5,
$$
which once again doesn't have any prime solutions (just need to check $q=2$ and $3$).