Let us call a number $N$ constructible if a regular $N$-gon can be constructed with straightedge and compass. As you know, this is equivalent to $\varphi(N)$ being a power of $2$, and equivalently $N$ is of the form $N=2^kF_{m_1}F_{m_2}\cdots F_{m_r}$ where $F_m=2^{2^m}+1$ is a Fermat number, the $m_i$ are all distinct and moreover each $F_{m_i}$ is prime. We interpret the product of the $F_{m_i}$'s as $1$ if $r=0$.
Let $b(N)$ be the number of $1$'s in the binary representation of $N$. Thus $b(15)=4$, $b(16)=1$. First of all, show:
1) $b(F_{m_1}\cdots F_{m_r}) = 2^r$.
This is so because when you expand the product $(2^{2^{m_1}}+1)(2^{2^{m_2}}+1)\cdots(2^{2^{m_r}}+1)$, the powers of $2$ don't "mix": Numbers of the form $2^{m_{i_1}}+2^{m_{i_2}}+\cdots+2^{m_{i_p}}$ are all distinct, so each term in the expanded product contributes precisely one $1$ to the binary representation of the result.
Now conclude:
2) If $N$ is even and both $N$ and $N+1$ are constructible, then $N=2^{2^m}$ for some $m$ and $N+1$ is a Fermat prime.
Proof: Since $N$ is even, $b(N+1)=b(N)+1$. By part 1) above we have $b(N)=2^r$ for some $r$ and $b(N+1)=2^s$ for some $s$. But the only way for $2^s=2^r+1$ to happen is if $r=0$ and $s=1$. Thus $N$ is a power of $2$ and $N+1$ is a constructible number of the form $F_m$, which necessitates that it's a Fermat prime and $N=2^{2^m}$.
We don't know how many $N$'s there are with this property, since we don't know how many Fermat primes exist. However, adding the requirement that $N-1$ is also constructible restricts the options enough for the problem to be solvable.
3) If $N$ is even and both $N-1$ and $N+1$ are constructible, then $N=2^{2^m}$ with $0 \leq m \leq 4$.
Proof: We already know that $N=2^{2^m}$ for some $m$. But now
$N-1 = 2^{2^m}-1 = F_0F_1\cdots F_{m-1}$ (proof: induction, or expand the product)
and for that to be constructible, each $F_k$ must be prime for $0 \leq k \leq m-1$. This works fine for $0 \leq m \leq 5$ but ceases to work afterwards since $F_5$ isn't prime, and for the same reason $m=5$ fails to ensure that $N+1$ is constructible. Thus the only possible values of $m$ are $0 \leq m \leq 4$.
Finally, we need to check triplets $(N-1,N,N+1)$ where $N$ is odd. This means that $N-1,N$ are now a power of 2 and a Fermat prime, respectively, with the additional requirement that $N+1$ is also constructible. I'll leave this as an exercise :-) The only possible cases are $N=3$ and $N=5$.
So finally, the list of constructible triplets is:
$(1,2,3)$, $(2,3,4)$, $(3,4,5)$, $(4,5,6)$, $(15,16,17)$, $(255,256,257)$, $(65535,65536,65537)$.