A possible Property of Euler's totient function: $n$ such that $\varphi(n)$ and $\varphi(n+1)$ are both powers of two $n$ is an odd positive integer such that $\varphi(n)$ and $\varphi(n+1)$ are both powers of two . Here , $\varphi(n)$ denotes Euler's totient function. 

Is it true that $(n+1)$ is either $6$ or a power of $2$? 

Please help me to prove or disprove this statement . (I was randomly flipping through some values attained by the $\varphi$ function when I observed this pattern) :)
 A: This is not an answer, but might be useful when someone smarter than me tries to prove this.
Write $n=\prod_{i}p_i^{n_i}$ and $n+1=\prod_{i}(p_i')^{m_i}$. Then $\phi(n)=\prod_{i} p_i^{n_i-1}(p_i-1)$ being a power of two implies that if $p_i\neq 2$, then $n_i=1$ and $(p_i-1)$ is a power of two. The same holds for $\phi(n+1)$.
Let $I$ be an enumeration of the prime numbers such that the prime number minus one is a power of two, i.e. $0$ corresponds to $2$, $1$ corresponds to $5$, $2$ corresponds to $17$ and so on. Then $$n=2^{n_0}\prod_{i\in I, i\geq 1}p_i^{n_i}$$ where $n_i\in \left\{0,1\right\}$ and $$n+1=2^{m_0}\prod_{i\in I,i\geq 1}p_i^{m_i}$$
where $m_i\in \left\{0,1\right\}$. Here $p_i$ denotes the $i$-th prime number such that $p_i-1$ is a power of $2$ (and we start counting from zero). From this we get that $$2^{n_0}\prod_{i\in I, i\geq 1}p_i^{n_i}+1=2^{m_0}\prod_{i\in I, i\geq 1}p_i^{m_i}.$$
From this equation, we should be able to find something, however, I do not even know whether the index set $I$ is finite or not, making further manipulations difficult.
