Why is $\gcd(x^4+1,x^2-1) = 1$ but I get $2$? [unit normalization of gcds]

If your polynomials are over $\mathbf{R}$, then a gcd of $1$ is equivalent to a gcd of $2$ (and any other nonzero real), because in general gcds are only well-defined up to associatedness, i.e. mutual divisibility.

The greatest common divisor of $f$ and $g$ is (in the case of polynomials) defined as a polynomial $d$ such that $d$ divides $f$ and $g$ and every divisor of $f$ and $g$ also divides $d$.

Thus, if we have another polynomial $e$ that is associated with $d$ (which means that $e|d$ and $d|e$), then we have $e|d|f$, $e|d|g$ and for every common divisor $z$ of $f$ and $g$ we have $z|d|e$, so $e$ is also a gcd of $f$ and $g$.

In your case you have two gcds of $1$ and $2$. Because $1\cdot 2 = 2$ and $2\cdot\frac12=1$ you have $1|2$ and $2|1$, so $1$ and $2$ are associated, thus if $1$ is a gcd, so is $2$ and vice versa.

Note that it is possible to normalize the gcd of polynomials by requiring the first nonzero coefficient to be $1$, which is what Wolfram|Alpha presumably does.

The general principle behind this is the concept of units in a ring (with a $1$-element): https://en.wikipedia.org/wiki/Unit_(ring_theory)

A unit is an element of the ring that divides $1$ (the multiplicative neutral element). When you are dealing only with multiplication (like in this case), elements that differ only by a unit have exactly the same behaviour regarding divisibility (I assume a commutative ring here):

If $u$ is a unit and $x=yu$, then $x$ and $y$ divide exactly the same ring elements:

$$x|a \iff \exists f:fx=a \Rightarrow (fu)y=a \Rightarrow y|a$$

and with $u'={1\over u}$ we have $y=xu'$ and we get similiarly

$$y|a \iff \exists f:fy=a \Rightarrow (fu')x=a \Rightarrow x|a.$$

One can show very similiarly, that exactly the same ring elements divide $x$ and $y$.

So to state this again, regarding divisibilty, $x$ and $y$ have exactly the same properties.

That means if you find by some calulation that $\gcd(a,b)=x$, then you can also say $\gcd(a,b)=y$ (always assuming that $x=yu$ with $u$ being a unit). That means the $\gcd$ is only determined up to a unit!

When starting on these topics of divisibility and $\gcd$ etc. with $\mathbb Z$, this is usually ignored, as the only units of $\mathbb Z$ are $1$ and $-1$, and the general presentation on these topics is centered on positive integers, so the $-1$ gets ignored.

To come back to your problem, the ring in questions is ${\mathbb Q[x]}$. It's easy to check that in that ring the units are exactly the non-zero constant polynomials. That means that any $\gcd$ is only determined up to a unit (=constant polynomial). It is just as valid to claim the $\gcd$ equals the polynomial $p(x)=1$ or $q(x)=15$ or $r(x)=-2019.0319$, or whatever other constant comes to your mind.

As the other answers said, using the polynomial with the leading coefficient equal to $1$ is a general convention so results can be easily compared. It's the same (a convention) as writing one half usually as ${1 \over 2}$ and not the equally valid ${-17 \over -34}$.

If you apply the Euclidean algorithm to $p(x)$ and $q(x)$ and the last non zero remainder is $44$, then there are polynomials $\alpha(x)$ and $\beta(x)$ such that $\alpha(x)p(x)+\beta(x)q(x)=44$. But then $\frac{\alpha(x)}{44}p(x)+\frac{\beta(x)}{44}q(x)=1$. So, $1$ is also a greatest common divisor of $p(x)$ and $q(x)$. Actually, if $r(x)$ is a greatest common divisor of $p(x)$ and $q(x)$, then so is $\frac{r(x)}k$, where $k$ is the coefficient of the monomial with highest exponent in $r(x)$. Then $\frac{r(x)}k$ is a monic polynomial, and it is usual to work with a monic polynomial when we are working with greatest common divisors of two polynomials.

Note that gcds are preserved under scaling by units (invertibles) because this holds true for divisibility, i.e. if $\,u\,$ is a unit then $\,ua\mid b\iff a\mid b\iff a\mid ub.\,$ Thus scaling a gcd by a unit does not alter its multiples nor its divisors, so it remains a gcd (i.e. a common divisor that is greatest divisibility-wise, i.e. divisible by every common divisor; said equivalently $\, c\mid a,b \!\iff\! c\mid \gcd(a,b),\, $ the universal definition of a gcd).

Often it is convenient scale gcds by a unit to a normal form, e.g. in $\,\Bbb Z\,$ we normalize gcds $\ge 0,\,$ and in a polynomial ring $\,K[x]\,$ over a field, we normalize them to be monic (lead coeff $\,c_n = 1),\,$ by scaling the polynomial by $\,c_n^{-1}\,$ if need be (thus a constant gcd $\,c_0\neq 0$ normalizes to $1).\,$ However, a nice unit normalization algorithm need not exist in all domains, so generally gcds are only determined up to unit multiples, i.e. up to associate-ness, so e.g. $\,\gcd(a,b)\approx 1$ means that the gcd is associate to $1$, i.e. is a unit, i.e $\,c\mid a,b\iff c\mid 1.\,$ Beware that a common abuse of language is to write $\gcd(a,b) = c\,$ to denote $\,\gcd(a,b)\approx c,\,$ esp. in contexts when there is no natural choice for unit normalization.

See here for further discussion.