Astronaut travels 16 lightyears and ages 15 years - am I misreading this question? 
The astronaut is travelling to a star sixteen lightyears away. During this trip he ages fifteen years. What is the speed u at which the astronaut travels?

It's obvious he can't be traveling at $\frac{16}{15}c$ (where $c$ is the speed of light), so I'm guessing this means we have to solve the equation:
$$16 = 15 \sqrt{1 - \frac{u^2}{c^2}}$$, with u the speed of the astronomer and c the speed of light in a vacuum.
Where I end up with $u = 0.37c$. Which is a lot more acceptable than the astronomer going faster than the speed of light. 
But then I'm assuming the time it takes him for an observer is equal to 16 years, am I correct in doing this? I think not since that means the astronomer would be travelling at the speed of light to an observer... 
 A: You got it half right, but you got so focused on the correction factor that you forgot to calculate the trip time in your rest frame.
In deference to Bill N, let me rephrase the question, hopefully more to his liking.
The astronaut is travelling to a star sixteen light years away which is stationary with respect to earth. During this trip he ages fifteen years - that is, an onboard clock records a trip duration of 15 years. What is the speed u, as measured from earth, at which the astronaut travels? Assume that the entire voyage is made at the velocity u (there is no acceleration or deceleration).
First, calculate the time T1 required for the trip in the earth, or rest, frame. This is simply $$T1 = 16\times{\frac{c}{u}} years$$ Now calculate the value of $\gamma$ which produces a duration T2 of 15 years in the moving ship $$T2 = \gamma \times T1 = \gamma\times 16 years \times \frac{c}{u} =15 years$$ So $$\gamma = \frac{15\times u}{16\times c} = 
\sqrt{1 - \frac{u^2}{c^2}}$$
You can solve for u/c, but by my figures$$\frac{u}{c} = 0.7295$$
Trip time in earth frame of reference is 21.93 years.
A: 
But then I'm assuming the time it takes him for an observer is equal to 16 years, am I correct in doing this? I think not since that means the astronomer would be travelling at the speed of light to an observer...

Okay, first thing: an astronaut (Greek, star sailor) is different from an astronomer (Greek, star namer). The former is seen at the wheel of a spaceship; the latter is seen at the wheel of a telescope.
Secondly: no, you should not assume that the time of flight is 16 years. Instead you have two statements that you need to simultaneously account for:


*

*There is a reference frame R1 which sees two "places" A and B (world-lines parallel to the time axis of R1) as having a constant minimum proper distance between them of $16\text{ ly}.$ 

*There is an astronaut travelling inertially from a point on world-line A to a point on world-line B. We call her reference frame R2. The proper time of her journey between these is $15 \text{ y}.$


The question is, how fast is she going? You are correct to intuit that these are likely reconcilable as long as the measurements are in different reference frames.
As for how to do this, probably the easiest way is to simply remember that for any two points in spacetime $(t_{0,1}, \vec r_{0,1})$, everybody agrees on the spacetime interval between them, $$I_{0\to 1} = c^2 (t_1 - t_0)^2 - |\vec r_1 - \vec r_0|^2.$$The information about R2 gives you the spacetime interval of her flight directly; she thinks that her spaceship is always "right here" so $|\vec r_1 - \vec r_0| = 0$ and the proper time she measures is therefore a direct measurement of a positive spacetime interval, $\sqrt{I} = 15\text{ ly}.$ That is a general thing about proper times: usually a proper time is a measurement of a spacetime interval.
Therefore in R1 where $D = 16 \text{ ly}$ we know that the actual elapsed time between these events is measured as $c t = \sqrt{I + D^2}$ and so $D/t = c / \sqrt{I/D^2 + 1}.$
A: No, you're not misreading the question.  From reference frame here on Earth, light would take 16 years to reach the start 16 light years away.
However, this problem takes a look at special relativity.  As the astronaut travels at relativistic speed, he experiences time dilation.  That means that clocks back on Earth will appear to tick fast as opposed to his clock on the ship.  As a result, the astronaut's clock on his ship only measures 15 years of elapsed time when he arrives at the distance star, whereas clocks in a rest frame measure at least16 years of elapsed time.
Please note that the amount of time it takes for light to travel a lightyear is measured from a rest frame.  So it is possible for him to travel $x$ lightyears and age less than $x$ years, as measured in his reference frame.
A: Based on the rewording well-given by @WhatRoughBeast we can consider two events: 


*

*the astronaut passes by Earth at $t_1 = t_1' = 0$ and $x_1=x_1'=0$, where the unprimed frame is at rest w.r.t. Earth, and the primed frame is at rest w.r.t. the astronaut.

*the astronaut arrives at the planet: $t_2=D/\beta$ and $x_2=D$ where $D$ is measured in light-time units (e.g., light-years) and $c=1$. In the frame of the astronaut, $t_2'=A\ \text{years}$ and $x_2'=0$ because the astronaut is at the location of the 2nd event.


In SR, the space-time interval for a pair of events is Lorentz invariant, so we can say (again, $c=1$)
$$\Delta t^2-\Delta x^2 = \Delta t'^2-\Delta x'^2$$
$$\frac{D^2}{\beta^2} - D^2 = A^2 - 0 $$
and finally
$$\beta^2=\frac{D^2}{A^2+D^2}$$
For the numbers in this particular problem, we get $\beta = 0.7295$.
We can also turn the problem around and ask, how much would the astronaut age in her ship if she travelled at $\beta=0.2$ with respect to Earth and the same planet.
$$0.2^2=\frac{16^2}{A^2+16^2}$$
or $A=78.4$ years. On Earth, 160 years have passed.
