= Solution
Regard
$$
F=X^2+Y^2-1
$$
as a monic quadratic in $X$ over $\mathbb Q[Y]$. By the <Gauss lemma for polynomials>, reducibility in $\mathbb Q[X,Y]$ would imply reducibility in $\mathbb Q(Y)[X]$, so $1-Y^2$ would be a square in $\mathbb Q(Y)$. This is impossible: its zero at $Y=1$ has odd order one, whereas every zero or pole of a square in a rational-function field has even order. Hence
$$
\boxed{X^2+Y^2-1\text{ is irreducible over }\mathbb Q}.
$$
The same <odd-valuation obstruction to a rational-function square> works over every field of characteristic different from two, since the zeros $Y=1$ and $Y=-1$ are then distinct. It does not work over every field. In characteristic two,
$$
X^2+Y^2-1=(X+Y+1)^2,
$$
so the <polynomial> is reducible.
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