Here's an elementary argument proving that $I$ is infinite.  

**Claim.** $p^{17} \notin I$ for all primes $p > 17$.  

**Proof.** Suppose that $p^{17}=\frac{N}{d(N)}$ for some $N$.  Write $N=p^{17+k}n$ where 17 does not divide $n$.  Then, $d(N)=(18+k)d(n)$, and so $p^kn=(18+k)d(n)$. Since $p>18$ and $n \geq d(n)$ this can only hold if $k=0$.  But now $\frac{n}{d(n)}=18$, which Greg Martin has shown is impossible.