I'll elevate my comment to an answer and give two more related ones. One seems less trivial for primes but has first exception at $30$, the other seems more obvious for primes but has first exception at $900$.

The cyclotomic polynomials $\Phi_d$ can be specified inductively by saying that, for all $n$, $\prod_{d|n}\Phi_d(x)=x^n-1.$ Equivalently, $\Phi_d(x)$ is the minimal polynomial of $e^{{2\pi i}/d}.$ It turns out that $\Phi_{15}=x^8-x^7+x^5-x^4+x^3-x+1.$ One might conjecture that the coefficients of $\Phi_m$ are always are always $0,1$ and $-1.$ This is true for primes, prime powers and even for numbers of the form $2^ip^jq^k$ (up to two distinct odd prime divisors) but it fails for $m=105$

The second example is of great interest to me, but takes a little explanation For a finite integer set $A$, we say that $A$ **tiles the integers by translation** if there is an integer set $C$ with $\{a+c \mid a \in A,c \in C \}=\mathbb{Z}$ and each $s \in \mathbb{Z}$ can be uniquely written in this form. Then we write $A \oplus C =\mathbb{Z}$. This property is not affected by translation so we will always assume that $0 \in A$ and $0 \in C.$

Consider this property enjoyed by certain integers $m$:

Whenever $A$ is an $m$ element set with $A \oplus C=\mathbb{Z}$ there is a prime divisor $p$ of $m$ such that $A \subset p\mathbb{Z}$ or $C \subset p\mathbb{Z}.$

It is true when $m$ is prime (but I don't consider it trivial) and also when $m$ is a prime power or a product $m=p^iq^j$ of two prime powers. It is not true for $m=30$ and other values with at least three distinct prime factors. The sets $A$ which provide counterexamples are rather spread out. If I recall correctly , a counterexample for $m=30$ will have $\max{A} \gt 720$ (if we set $\min{A}=0$. )

Here is a variant form: Write $A \oplus B=\mathbb{Z}_n$ when $A \oplus B$ is a complete set of residues $\mod n=|A||B|.$ Here we will assume $0=\min{A}=\min{B}$ and consider this property which is enjoyed by certain integers $n$.

Whenever $A \oplus B=\mathbb{Z}_n$ , there is a prime divisor $p$ of $n$ such that $A \subset p\mathbb{Z}$ or $B \subset p\mathbb{Z}.$

It always holds when $n$ is a prime, or prime power or even a product of two prime powers $n=p^jq^k.$ It fails when both $|A|$ and $|B|$ can have three distinct prime divisors so the first time is for $n=2^23^25^2=900$ as well as for $n=2\cdot3\cdot5\cdot 7 \cdot 11 \cdot 13=30030.$ So, while this seems trivial as a property of $n=|A||B|$, it is actually a property of $\min(|A|,|B|)$ (although it would take longer to explain why) and is not trivial when that minimum is a prime.

Now that I got to the property resisting digressions, let me explain why it is interesting (optional), mention the existence of an open problem and demystify the property a bit. For details see Tiling the integers with translates of one finite set which also proves the claims above and shows a link to cyclotomic polynomials.

It is interesting to characterize finite sets $A$ which tile the integers by translation: $A \oplus C=\mathbb{Z}.$ There are attractive sufficient conditions (T1 and T2 in the linked paper). These conditions are necessary when the size has at most two prime divisors, $|A|=p^{\alpha}q^{\beta}.$ The method of proof depends strongly on the property above. It is also not hard to show that if $A \subset p\mathbb{Z}$ (all elements of $A$ are multiples of $p$) Then there is $C$ with $A \oplus C=\mathbb{Z}$ if and only if there is a set $C'$ with $A' \oplus C'=\mathbb{Z}$ `where $A'=\lbrace\frac{a}{p} \mid a \in A \rbrace.$`

This reduction to a smaller case (along with the rest and a bit more) is what allows the proof that the sufficient conditions are also necessary for a set of size $|A|=p^{\alpha}q^{\beta}$ to tile the integers by translation. It is possible that the conditions are necessary for $A$ of any finite size, however the method of proof would have to be quite different. The first potential exception would be for $A$ with $30$ elements which tiles $\mathbb{Z}_{900}.$

Here is a way to restate the property above so that it does hold for all $m$ (but fails in general to allow the proof of necessity): If $A \oplus B=\mathbb{Z}_n$ then for one of the two sets , say $A$, none of the differences $a_i-a_j$ is coprime to $n$. Since $0 \in A$ this means that also every $a \in A$ shares a divisor with $n.$ So if $n=72$ then every member of $A$ and every difference is divisible by $2$ or $3$ or both. In fact they are all even or all multiples of 3 lest there be $a_x \in A$ not divisible by $2$ and $a_y \in A$ not divisible by $3$ as then $a_x-a_y$ would share no prime divisors with $72.$ So the reduction is possible and a theorem can be proved. When $A$ has 30 elements it can be the case that among them are $6,10,15$ and various of their multiples so the elements and differences all share a divisor with $30$ but no one divisor covers all cases and the proof is not available.