Number of prime divisors with multiplicity in a sum of Gaussian binomial coefficients

Question

For $q$ a prime and $t \geq 0$ let $a_t^q=\sum\limits_{k=0}^{t}{[t,k]_q}$ with $[t,k]_q$ the Gaussian binomial coefficient, see https://en.wikipedia.org/wiki/Gaussian_binomial_coefficient.

Question: For which $t \geq 1$ and $q$, do we have that $a_t^q+1$ has at least $t+1$ prime divisors counted with multiplicity and additionally $a_t^q+1 \geq q^{t+1}$?

I have not yet found an example.

In case I made no mistake (but it is late here...), an example of such $t$ and $q$ would give examples answering a question of Jeremy Rickard in The number of ideals in a ring . Namely, the ring $R=R_n^q=K[x_1,...,x_n]/\langle x_1,...,x_n\rangle^2$ with $K$ a finite field with $q$ elements has $q^{n+1}$ elements and thus |R| has $n+1$ prime divisors with multiplicity. The number of ideals of $R$ should be $a_n^q+1$.

Easy example with $n=1$: $R_1^q=F_q[x]/(x^2)$ has the 3 ideals $0,(x)$ and $R_1^q$ and the ring has $q^2$ elements.

Answer 1

Here is an example (I hope I did this right!): With $t=5$ I get $a_5^q=2 q^6+2 q^5+6 q^4+6 q^3+6 q^2+4 q+6$. In particular $a_5^q+1\geq q^{5+1}$ for any prime $q$. Letting $q=79$ we get $a_5^q+1 = 492\,565\,720\,329 = 3^3\cdot 17\cdot 59\cdot 107\cdot 169987$ which has $7$ prime factors counted with multiplicity.