jmc

We could find the answer by trial and error -- testing each candidate $t$ to see if $t\cdot (t+2)\equiv 1(\mathrm{mod}23)$. However, here is another way:

We can easily see that $4\cdot 6=24\equiv 1(\mathrm{mod}23)$, so $4$ fulfills the main requirement that its inverse is $2$ more than it. Unfortunately, $4$ isn't odd. But we also have $$\begin{array}{rl}(-4)\cdot (-6)& =4\cdot 6\\ & \equiv 1(\mathrm{mod}23),\end{array}$$ so $-4$ and $-6$ are each other's inverses $(\mathrm{mod}23)$. Since $-4\equiv 19(\mathrm{mod}23)$ and $-6\equiv 17(\mathrm{mod}23)$, the answer $t=\boxed{17}$ satisfies the requirements of the problem.

(We can even check that $17\cdot 19=323=14\cdot 23+1$.)