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As a continuation of this question I would like to ask about possible orders of trace $k$ elements in $SL_2(q)$. Here are examples which I know.

When trace is zero then we have $x^2=-1$ so it means that order of $x$ is $2$ in characteristic $2$ and it is $4$ in odd characteristic.

When trace is $-1$ then we have $x^2=-x+1$, so $x^3=-x^2+x=-x+1+x=1$. It means that order of $x$ is $3$.

When trace is $t$ which is order $q-1$ element in case $q=2^n$. Then order of $x$ is sometimes $q+1$ and $x$ generate $\mathbb F_{q^2}$ subalgebra. This is just guess, I have only checked this for $q=2,4,8,16$. Here is small test in GAP showing order of trace $t$ elements for $q=2^n$ for $n=1..10$, where $t$ is generator of the field multiplicative group:

gap> List([1..10],k->Order([[Z(2^k),1],[1,0]]*Z(2^k)^0));
[ 3, 5, 9, 17, 31, 65, 43, 51, 511, 25 ]

Let's call element imaginary when it is of trace $0$ in $SL_2(q)$. From above we know that order of such element is either $2$ or $4$. The next question we can ask is what order can have product of two imaginary elements. According to tests in GAP in characteristic two we obtain orders $q-1$, $q+1$ and divisors which are all orders in the group (tested only for few small q). In case of odd characteristic I do not have theory ready yet. Anyway set $\{x^2=-1\}$ seems to be interesting.

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Here is some experimental data from GAP for answering this question. I do not have full picture yet. Let $q=p^n$, $p$ is prime number.

There are three types of subalgebras with one generated by one element: $\mathbb F_q+\mathbb F_q$, $\mathbb F_q+\mathbb F_q\pmb i$, $\mathbb F_{q^2}$ where $\pmb i^p=1$. The groups contained with invertible elements are $C_{q-1}\times C_{q-1}$, $C_{2(q-1)}\times \underbrace{C_p\times...\times C_p}_{n-1}$, $C_{q^2-1}$ with sizes $(q-1)^2,q^2-q,q^2-1$ respectively.

The three cases are distinguished by order of generator $u$ of the subalgebra. When it is divisor of $q-1$ then we are in case 1. When it is divisor of $q$ then we are in case 2. When it is divisor of $q+1$ then we are in case 3. Common divisor of $q-1$ and $q+1$ is $2$ and it happens in odd characteristic. In this case element of order $2$ generate case 1. In characteristic 2 zero divisor $p^2=p$ generate subalgebra of type 1. It is element of $Q_{01}$ (see this question for notation) with trace equal $1$.

Element of order $p$ in odd characteristic is belonging to $Q_{12}$ i.e. it is element of determinant $1$ and trace $2$ (and not belonging to $\mathbb F_q1)$.

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