$m$ is beautiful iff $4m=k^2+3$ where $k$ is odd. Now there are inf many odd primes $p$ such that $-3$ is a square mod $p$, this follows from QR. Then $k^2\equiv -3(p^2)$ also has a solution by Hensel's lemma. Replacing $k$ by $k+p^2$ if necessary we may assume that $k$ is odd.
Now for each odd prime as above we find a beautiful $m_p$ such that $p^2|m_p$. Then the set $\{m_p\}$ must be infinite and we are done.
@parmenides: I understand now, what they mean is "and such that the square parts of different beautiful integers are relatively prime". Ευχαριστω για την εξηγηση σου και γενικα για τα ωραια προβληματα που μας δινεις.
