Find all integers $n>1$ with the property that for each positive divisor $d$ of $n$, we also have that: $$ (d+2) \mid(n+2) $$
So far, I've tried to get as much info about it as possible, but I don't know how to proceed.
My attempt so far: First, we know that $n$ is of the form of the form $3K+1$. Also, if you take the smallest prime divisor of $n$, call it $p$, we know that:
$$ n=pd \Rightarrow d+2 \mid pd+2 \Rightarrow d+2 \mid 2p-2 $$ Hence $p \leq d \leq 2p-4$.
You've made a good start, with this showing how to complete the solution by checking the multiplier. First, with $d = 1$ always being a divisor, we require that $1 + 2 \mid n + 2 \;\to\; n \equiv 1 \pmod{3} \;\to\; n = 3k + 1$, as you've already stated. Since the only other factor of primes is the prime itself, and $d = n$ always satisfies $d + 2 \mid n + 2$, this means that all primes $n$ where
$$n \equiv 1 \pmod{3}\tag{1} \label{eq1A}$$
are solutions. Otherwise, similar to what you did, for any composite $n$, have $p$ be its smallest prime factor. Then $d = \frac{n}{p}$ is also a factor of $n$ so, for some positive integer multiplier $m$, we have
$$\begin{equation}\begin{aligned} n + 2 & = m\left(\frac{n}{p} + 2\right) \\ n + 2 & = \frac{mn}{p} + 2m \\ pn + 2p & = mn + 2pm \\ (p - m)n & = 2p(m - 1) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note from the second line above that $m \lt p$ since, otherwise, the RHS is greater than the LHS. Thus, if $p = 2$, then $m \lt 2 \;\to\; m = 1$, but then the last line becomes the incorrect $n = 0$. Otherwise, with $p \gt 2$, then $2 \nmid n$. The last line then shows that $2 \mid p - m$. However, we then get that $n \le p(m - 1) \le p(p - 2)$, but since $n \ge p^2$ (as $n$ is composite and $p$ is the smallest prime factor), this is also not possible.
This proves the only solutions are those given in \eqref{eq1A}, i.e., $n$ which are primes of the form $3k + 1$.