Exercise 2.4.11 of Qing Liu's `Algebraic Geometry and Arithmetic Curves'
Let A be a ring with an ideal $I$. If $$f : X \to \mathrm{Spec}(A)$$ is a quasi-compact morphism, then $$f(X) \subset V(I) \iff f^{\#}(\mathrm{Spec}(A))(I) \subset \sqrt{0}.$$
I've managed to show $\impliedby$ by taking an affine cover $(\mathrm{Spec}(A_{i}))$ of $X$, and showing that, since $f^{\#}(\mathrm{Spec}(A))(I)$ is nilpotent then the image under the restriction maps is nilpotent, $I$ is contained in the inverse image of all of the prime ideals of $A_{i}$ i.e. $f(x) \subset V(I)$ for all $x \in \mathrm{Spec}(A_{i})$ for all $i$ and thus for all $x \in X$.
The if and only if for $X$ affine is easy, but I'm having trouble using the quasi-compactness to prove it for general $X$. Any hints?
By quasi-compactness of $f$, $X$ is quasi-compact.
Since you have done the affine case, you have already shown:
Any element $a \in f^{\#}(\mathrm{Spec}(A))(I)$ is nilpotent after restriction to an affine open.
Since $X$ can be covered by finitely many affine opens, we find a common nilpotency index, say $a^n$ is zero on all affine opens of the cover. Of course $a^n=0$ globally follows.
Basically we have shown that locally nilpotent and globally nilpotent are the same for quasi-compact schemes. If $X$ is not quasi-compact, locally nilpotent and globally nilpotent are not the same notions. That is why you need quasi-compactness.