Where should the exponent be written in numbers with units of measurement? If you are to calculate the hypotenuse of a triangle, the formula is:
$h = \sqrt{x^2 + y^2}$
If you don't have any units for the numbers, replacing x and y is pretty straightforward:
$h = \sqrt{4^2 + 6^2}$
But what if the numbers are in meters?
$h = \sqrt{4^2m + 6^2m}$ (wrong, would become $\sqrt{52m}$)
$h = \sqrt{4m^2 + 6m^2}$ (wrong, would become $\sqrt{10m^2}$)
$h = \sqrt{(4m)^2 + (6m)^2}$ (correct, would become $\sqrt{52m^2}$)
Or should I just ignore the unit of measurement in these cases?

Suppose you have been given $x$ and $y$ in metres, and you'd like to know the quantity, $z=\sqrt{x^2+y^2}$. Then, as you have predicted this quatity will be in metres. 
Two things have been involved:
Homogeneity of Dimension
Two quantities of different dimensions cannot be added. This is one of the axioms of numerical physics. 
Example: It is clear that adding $5$ metres to $3$ seconds does not give a physically meaningful quantity that can be interpreted in real life.
Certain functions only take values in dimensionless quantities
For instance, $\sin (\sqrt{x^2+y^2})$ would not make sense even if $x$ and $y$ have same dimensions. This is a bit subtler, but this is what it is!



*

*Coming to your question, the first quantity you tell us in dimension of $m^{1/2}$ which is against your guess!

*The second quantity is dimensionally fine while numerically this is not what you want.

*The third quantity is fine in all ways.

My  suggestion:
First manipulate the numbers and then the units separately. This is a good practice in Numerical Physics. The other answers have done it all at one go. But, I don't prefer it that way!
