Find all integers $n$ that are coprime with every number of the form $2^{m} + 3^{m} + 6^{m} - 1, m \in \mathbb{N}$.
We will show that only 1 is suitable. For this, it suffices to establish that for any prime $p$, there exists an integer $m$ such that $p \mid 2^{m}+3^{m}+6^{m}-1$. For $p=2$, any $m$ works. For $p=3$, any even $m$ (for example, 2) works. For $p \geq 5, p$ is coprime with $2,3,6$. We leave it to the reader to study the cases $p=5, p=7, p=11$ (by calculating the residues of $2^{m}+3^{m}+6^{m}-1$ modulo $p$) and note that in these cases $p \mid 2^{p-2}+3^{p-2}+6^{p-2}-1$. Let's show that this result is general.

For $p$ prime greater than $5,2^{p-2}+3^{p-2}+6^{p-2} \equiv 6^{p-2} * 3 * 2 * 2^{p-2}+6^{p-2} * 2 * 3 * 3^{p-2}+6^{p-2} \pmod{p}$, according to Fermat's Little Theorem, and the second term is congruent modulo $p$ (still by Fermat's Little Theorem) to $6^{p-2}(3 * 1+2 * 1+1) \equiv 6 * 6^{p-2} \equiv 1 \pmod{p}$, which is what we wanted to prove and concludes the proof.