 Solution

The strict inequality means $v_3(x^2+2)>2$. Taking $x=5$ gives $x^2+2=27$, soFor the other expression, if $3\nmid x$ then $x^2+3\not\equiv 0(\mathrm{mod}3)$ and its P-adic valuation is zero. If $3\mid x$, then $x^2+3=3(3(x/3{)}^2+1)$ has P-adic valuation exactly one. It can never have valuation greater than two. Thus no such integer exists for the second strict inequality.