Let me answer your question as best I can; but I'll start by correcting a misconception in your question.
It actually never happens that "all units are powers of just one unit" except when the unit group is finite. The reason this happens is because $-1$ gets in the way! Suppose there's a unit $u$ such that every other unit $v$ can be written as $u^n$ for some $n \in \mathbf{Z}$. But $v = -1$ is a perfectly good unit, so we must be able to write $u^n = -1$; so this equation forces $u$ to be a root of unity, and in particular it can't generate an infinite group.
When the unit group is infinite, the best we can hope for is that there is a unit $u$ such that every unit $v$ is $u^n z$ where $z$ is a root of unity. This is precisely what happens for real quadratic fields: the fundamental unit $u$ has the property that every unit is $\pm u^n$ for some $n \in \mathbf{Z}$ (and you can't get rid of the $\pm 1$ there).
When $K$ has bigger degree you need more than one unit, and Dirichlet's unit theorem gives a precise formula for how many units you need. If the number field $K$ is generated by a single algebraic number $\theta$, then you look at the minimal polynomial of $\theta$ and count how many real and non-real roots it has: it will have $r$ real roots and $s$ conjugate pairs of non-real roots, for some $r$ and $s$. Dirichlet's theorem says that the smallest collection of units you need to get within a root of unity of every unit in $K$ is of size $r + s - 1$.
For instance, I'm particularly fond of the field $K = \mathbf{Q}(\sqrt[3]{2})$. The minimal polynomial of $\sqrt[3]{2}$ is $X^3 - 2$ which has one real root and two conjugate non-real roots; so in this case one unit suffices and the field $K$ does have a fundamental unit, which turns out to be $\sqrt[3]{2} - 1$.
On the other hand, the field $L = \mathbf{Q}(\sqrt[4]{2})$ corresponds to $X^4 - 2$, which has two real roots and one pair of conjugate roots; so we'll need $2 + 1 - 1 = 2$ units to generate everything -- there's no "fundamental unit" for $L$. There are computer programs for calculating in number fields, and my computer took approximately 0.07 seconds to tell me that if we take $u_1 = \sqrt[4]{2} + 1$ and $u_2 = \sqrt{2} - 1$ then every unit of $L$ is of the form $\pm u_1^a u_2^b$ for some $a, b \in \mathbf{Z}$.
PS: I will try and answer your updated questions.
You ask whether it wouldn't be better to take $\{ \sqrt[4]{2} + 1, \sqrt[4]{2} - 1\}$ rather than $\{ \sqrt[4]{2} + 1, \sqrt{2} - 1\}$ as "fundamental units". This is entirely a matter of taste: there's no really 'best' set to take, and lots of sets will work equally well. When $r + s -1$ is 1, then a fundamental $u$ will be unique up to replacing $u$ with $\omega \cdot u^{\pm 1}$ for $\omega$ one of the finite set of roots of unity in the field (usually just $\pm 1$); but as soon as you step to larger degrees there'll be an infinite amount of choice and it is a rather fruitless exercise to try and single out any one choice which is 'best'.
As for actually computing units in practice: the bible for such computations is Henri Cohen's book "A Course in Computational Algebraic Number Theory".
Here’s the story, though I leave it to others to furnish a proof or give a proper reference.
For $d\ge2$ and not a square, you get all solutions to the Pell equation $m^2-dn^2=\pm1$ by looking at the continued fraction expansion of $\sqrt d$. If $k=\lfloor\sqrt d\rfloor$, then your expansion looks like this:
$$
\sqrt d=k+\bigl[\frac1{\delta_1+}\>\frac1{\delta_2+}\cdots\frac1{\delta_{r-2}+}\>\frac1{\delta_{r-1}+}\>\frac1{2k+}\>\bigr]\,,
$$
where the part in brackets repeats infinitely. Then, every time you evaluate a convergent to the continued fraction just before the appearance of the $2k$, you’ll get a solution of Pell from the numerator $m$ and the denominator $n$. For instance, $\sqrt7=2+\frac1{1+}\,\frac1{1+}\,\frac1{1+}\,\frac1{4+}\,\cdots$, repeating with a period of length four. You evaluate $2+\frac1{1+}\,\frac1{1+}\,\frac1{1}=8/3$, and lo and behold, the first solution of $m^2-7n^2=\pm1$ is $m=8$, $n=3$.
For $d$ squarefree and incongruent to $1$ modulo $4$, this gives you all the units in $\Bbb Q(\sqrt{d}\,)$, but for $d\equiv1\pmod4$, it may happen that what you get is the cube of a unit. But as I recall, in that case, the primitive unit’s coordinates always will show up as the numerator and denominator of an earlier convergent.
Best Answer
The usual method is factoring elements of small norm. In your case, the ideals generated by $\alpha = 8 + \sqrt{61}$ and $\beta = (7 - \sqrt{61})/2$ have norm $3$, and the quotient is a nontrivial unit, namely $\frac{39 - 5 \sqrt{61}}2$.
Afterwards you have to show that $\frac{39 + 5 \sqrt{61}}2$ is not a power, which is not hard since the fundamental unit is $\ge \frac{1 + \sqrt{61}}2$, and this bounds possible exponents.