If one imagines that $f(a) \bmod n^2$ is about uniformly distributed (for most large enough $n$ ) then the probability of having a small root $\bmod n^2$ is $1/n$ so that would point to infinitely many small roots but very sparsely distributed.

At least up to $n=10000$ there are no small roots $\bmod n^2$ for $f(x)=x^3+2$.

**LATER** $f(34697)= 53\cdot239\cdot57425^2$

An indication that maybe things are not so random, at least for this polynomial, are these interesting cases with $\frac{a}{n}$ reasonably small.

$f(208)=6\cdot131\cdot107^2$ with $\frac{a}{n} \approx 1.944.$

$f(224)=22\cdot43\cdot109^2$ with $\frac{a}{n} \approx 2.055.$

$f(3440)=2\cdot27329\cdot863^2$ with $\frac{a}{n} \approx 3.986.$

$f(3472)=6\cdot9323\cdot865^2$ with $\frac{a}{n} \approx 4.014.$

$f(17472)=2\cdot313848\cdot2915^2$ with $\frac{a}{n} \approx 5.99383.$

$f(17520)=2\cdot251\cdot1259\cdot2917^2$ with $\frac{a}{n} \approx 6.00617.$

These three pairs have some remarkable properties which I can't explain. Among them are

$107,109=2^23^3\pm1$ and $208,224=2^33^3\pm1\cdot2^3$

$863,864=2^53^3\pm1$ and $3440,3472=2^73^3\pm2\cdot2^3$

$2915,2917=2^73^6\pm1$ and $17472,17520=2^33^7\pm3\cdot2^3$

One other case not in a pair like this is

$f(2272)=6\cdot1019\cdot1385^2$ with $\frac{a}{n} \approx 1.64.$

It is true that $2272=2^83^2-8\cdot2^3$ but $f(2^83^2+8\cdot2^3)$ is square free.

**LATER**

The previous results (except $(a,n)=(34697,57425))$ were unintelligent brute force. Slightly more intelligent is the fact that there is code to solve $f(x) \mod m$ (so here $m=n^2$.) Using that I found the mentioned solution and, as expected, a fourth and fifth pair with $\frac{a}{n}$ very close to $8$ and to $10$.

The $(a,n)$ pairs are

$(2^83^3\pm1,2^{11}3^3\pm4\cdot2^3)$

$(2^23^35^3\pm1,2^33^35^4\pm2^35^1)$