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Is a scheme being Noetherian equivalent to the underlying topological space being Noetherian and all its stalks being Noetherian?

This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point":

Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $alpha in k$ let $R_alpha = R[y_alpha]/((x-alpha)y_alpha,y_alpha^2)$. Then $R_alpha$ is an affine line with an embedded prime $mathfrak p_alpha = (x-alpha,y_alpha)$ at $x = alpha$, sticking out in the $y_alpha$-direction. Finally, let
$$R_infty = bigotimes_{alpha in k} R_alpha = operatorname*{colim}_{substack{longrightarrow\I subseteq k\text{finite}}} bigotimes_{alpha in I} R_alpha$$
be their tensor product over $R$ (not over $k$); that is
$$R_infty = frac{k[x]left[{y_alpha}_{alpha in k}right]}{sum_{alpha in k}((x-alpha)y_alpha, y_alpha^2)}.$$
This is not a Noetherian ring, because the radical $mathfrak r = ({y_alpha}_{alpha in k})$ is not finitely generated. But $operatorname{Spec} R_infty$ agrees as a topological space with $operatorname{Spec} R_infty^{operatorname{red}} = mathbb A^1_k$, hence $|!operatorname{Spec} R_infty|$ is a Noetherian topological space.

On the other hand, the map $R to R_alpha$ is an isomorphism away from $alpha$, and similarly $R_alpha to R_infty$ induces isomorphisms on the stalks at $alpha$. Thus, the stalk $(R_infty)_{mathfrak q_alpha} = (R_alpha)_{mathfrak p_alpha}$ at $mathfrak q_alpha = mathfrak p_alpha R_infty + mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $mathfrak r$ is just $R_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R_infty$ are Noetherian. $square$

Remark. As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I_0 subseteq I_1 ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$. Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$. Then you can try to use the Noetherian rings $mathcal O_{X,x}$ for the generic points $x$ of $V$.

However this is not quite true, because of embedded points. The precise statement [Tags 0311 and 02M3] is that the inclusion $I_{i-1} subseteq I_i$ is an equality if and only if the same holds for the localisation at every associated point of $I_i$. Now you run into trouble if the $I_i$ differ at an embedded point that is not seen by the topology of $V$. Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $mathcal O_{X,x}$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.

A further counterexample is Example 2.3 in W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.