Non-commutative finite division rings (or rather not)

Let's skip characteristic $2$ for the moment, so I'll assume $F$ is a field with characteristic $\ne2$.

The $F$-algebra $H(F)$ of quaternions over $F$ that you define has the antiautomorphism $q\mapsto q^*$, where, for $q=w+xi+yj+zk$, $q^*=w-xi-yj-yk$. It is bijective and preserves addition, but satisfies $$ (q_1q_2)^*=q_2^*q_1^* $$ by direct computation.

If we define $N(q)=qq^*=q^*q$, we see that $N(q)\in F$. Now, if $N(q)\ne0$, the quaternion $q$ is invertible, because $$ \frac{q^*}{N(q)}q=1=q\frac{q^*}{N(q)} $$ On the other hand, if $N(q)=0$, the quaternion $q$ is a zero-divisor by definition.

Suppose $H(F)$ is not a division ring. Take a nonzero noninvertible quaternion $q=w+xi+yj+zk$; then $N(q)=w^2+x^2+y^2+z^2=0$ and therefore we can write $-1$ as a sum of squares in $F$.

According to Artin-Schreier theory of formally real fields, a field where $-1$ is not a sum of squares can be ordered, so it has characteristic $0$ and is infinite.

Over a field of odd prime characteristic, it may happen that $-1$ is a sum of squares, but you need more than three. However, the four-square theorem you mention proves that the quaternion algebra is not a division ring also in this case: indeed, if the characteristic is $p$, then $p=a^2+b^2+c^2+d^2$ for some integers $a,b,c,d$ and therefore the quaternion $q=a+bi+cj+dk$ (where $a,b,c,d$ are interpreted in $F$, as usual) has $N(q)=0$.

By the way, in Herstein's “Topics in Algebra”, Wedderburn's theorem is exploited to prove the four-square theorem.

What happens in characteristic $2$? Well, the conjugation is the identity, so it's not really useful. However, the quaternion $1+i$ is not invertible, because $(1+i)^2=0$.

In conclusion, $H(F)$ is never a division ring, unless $F$ has characteristic $0$ and $-1$ cannot be written as a sum of at most three squares. Finiteness of $F$ is not needed.