Find $q$ such that $[q[qn]]+1=[q^2n]$ for $n=1,2,\dotsc$

First, let's reduce this to a question about fractional parts. First we rewrite the equation as follows: $$[q^2n-q\{qn\}]+1=q^2n-\{q^2n\}$$ Or, equivalently, $$q^2n-\{q^2n\}-1 \leq q^2n-q\{qn\}<q^2n-\{q^2n\}$$ Then, making cancellations, we reduce to $$\{q^2n\}+1 \geq q\{qn\} >\{q^2n\}.$$ Now, we can try to get into trouble: The reason this works for the golden ratio is that $\{\varphi^2n\}=\{\varphi n\}$, which makes this simpler. Note also that, if $q$ is rational written as $q=a/b$, then setting $n=b^2$ violates right inequality, so $q$ must be irrational. Then, assuming $q$ is irrational, our condition is equivalent to the following (closed) condition: $$\{q^2n\}+1 \geq q\{qn\} \geq \{q^2n\}.$$

Note now, all that matters is the set $S=\{(\{q^2n\},\{qn\}):n\in\mathbb N\}$ and $q$ itself. We require the following hold for all $(x,y)\in S$: $$x+1 \geq qy\geq x.$$ Now, note that if we define $\bar S$ as the closure of $S$, this condition must still hold on $\bar S$.

It turns out - from either casework or knowing something about diophantine approximation, observing that $q$ is irrational - that $\bar S$ will always either be a set of the form $\{(\{\alpha\},\{c\alpha\}):\alpha\in \mathbb R\}$, where $c$ is rational (and where we take the closure of this set in $[0,1]\times [0,1]$) or must be all of $[0,1]\times [0,1]$.

Observe first that $(1,1)\in \bar S$ in any case, so we get that $2\geq q\geq 1$, where we can make these inequalities strict since $q$ is irrational. Then, observe that if $(x,0)\in \bar S$, then $x=0$, since otherwise the condition would be violated. Then, suppose that $(1,y)$ is in $\bar S$; we get here that $qy\geq 1$. Since $q<2$, this implies $y>1/2$, but unless $y=1$, we would also get that $(1,1-y)$ is in $\bar S$ from looking at the possible cases for $\bar S$ - but $1-y<1/2$. Thus, it must be that $y=1$ in this case.

Therefore, we have established that the intersection of $\bar S$ with the boundary of the square $[0,1]\times [0,1]$ is just $(0,0)$ and $(1,1)$. In particular, this analysis gives us that $\bar S=\{(x,x):0\leq x \leq 1\}$. This can only happen if $\{q^2\}=\{q\}$. Then, finally, this occurs for solutions to $q^2-q-n$. The solution to this is $q=\frac{1\pm\sqrt{1+4n}}{2}$. Then, the only solution for $q$ in the interval $(1,2)$ is $q=\varphi$, so this is the only number with this property.