Rotman, Corollary 5.45. Rotman's book on group theory includes the following proposition, as a corollary of the fact that $\Bbb Z_{2^n}^\times \simeq C_2\times C_{2^{n-2}}$

Corollary 5.45 Let $G$ be a group containing elements $x,y$ with $|x|=2^m$, $y^2=x^{2^r}$ and $yxy^{-1}=x^t$. Then $t=\pm 1,\pm 1+2^{m-1}$, and in this last case $G$ contains at least two involutions.

You can see his proof here: http://librarum.org/book/1636/141
But note something is wrong. First, one can see the typo $$k2^{m-1}+2^r=2^r(k2^{m-r-1}\color{red}{-}1)$$
However, there is a wrong claim, namely that we can solve $k2^{m-r-1}=-1\mod 2^{m-r}$ for $k$. Of course this is possible if $r+1=m$, by $k=1$, but if $m>r+1$ this is impossible: if $k$ is even, we get $0=-1\mod 2^{m-r}$, and if $k$ is odd it can simply be cancelled to give $2^{m-r-1}+1=0\mod 2^{m-r}$, again impossible.

The corollary has the following mistake: if $t=2^{m-1}-1$, the congruence $k2^{m-r-1}=-1\mod 2^{m-r}$ is not solvable for $k$ if $m\neq r+1$: if $k$ is even, one gets $0=-1\mod 2^{m-r}$, and if $k$ is odd one can simply cancel it to get $2^{m-r-1}+1$, an odd number, divisible by an even number. If $m=r+1$; $x^{-1}y$ is an involution. I prove that if $t=-1+2^{m-1}$, then this is always the case, so there is always an involution, and the only case possible in your conclusion of theorem 5.46, assuming $G$ is non-abelian, is $t=-1$. The book has the typo $k2^{m-1}+2^r=2^r(k2^{m-r-1}{\color{red}-}1)$. Here is the proposed correction. 

$ \bf Corollary 5.45$ Let $G$ be a group containing elements $x,y$ such that $x$ has order $2^m$, $m\geqslant 3$, $y^2=x^{2^{r}}$ and $yxy^{-1}=x^t$. Assume further that $y\notin \langle x\rangle$. Then $t=\pm 1,\pm 1+2^{m-1}$. If $t=1+2^{m-1}$, $G$ has more than one involution for any choice of $r$. If $t=-1+2^{m-1}$, we must have $r\geqslant m-1$ and hence $r=m-1$, since we may assume $r\leqslant m-1$. In this case, $x^{-1}y$ is a second involution.

$\bf Proof$ I will only supply the correction to the claim there is an involution for $t=-1+2^{m-1}$. If this is the case, then $yxy^{-1}=x^{-1+2^{m-1}}$ gives $yx^2y^{-1}=x^{-2+2^m}=x^{-2}$, hence $x^{2^{r}}=yx^{2^r}y^{-1}=x^{-2^r}$. Thus $2^{r+1}=0\mod 2^m$; hence $m\leqslant r+1$. The rest of the claim is trivial verifications, and of course the repeated use that $y$ is not a power of $x$. $\blacktriangleleft$
$\bf Comment$ Observe that this avoids any modification to the proof of theorem 5.46: if $t=\pm 1+2^{m-1}$, we must have involutions, and since $G$ is non-abelian, we discard $t=1$, leaving $t=-1$ our only option, and concluding $G$ is isomorphic to $Q_{2^n}$.
