differential-privacy: show $\epsilon$ -differentially privacy

In this problem we consider a sensitive dataset $$x \in \{−1, 1\}^n$$. We consider the bounded setting where neighboring n-dimensional datasets differ in one coordinate. $$A$$ mechanism is available that supports statistical queries on x. Specifically, for a query $$q \in \{−1, 1\}^n$$ the mechanism computes the dot product $$\langle x, q \rangle = \sum_{i=1}^n x_iq_i$$ and releases $$\langle x, q \rangle + z$$ where $$z \sim Lap(\lambda)$$ is Laplace-distributed noise added to protect the privacy of individual entries of x.

My thoughts are that the $$\langle x,q \rangle$$ will differ between $$-1$$ and $$1$$, so we have $$\{-1,1\}^n+z$$ where $$z \sim Lap(\lambda)$$. Therefore when considering two databases $$x_1, ..., x_n$$ and $$x_1, ..., x'_n$$, we get that: $$\begin{array}[ccc] \;\frac{P(m(x_1, \dots, x_n) = t)}{P(m(x_1, \dots, x'_n) = t)} & \leq & \exp(2/\lambda)\\ & = &\exp(\varepsilon) \end{array}$$ But I'm not sure if it correct? Especially not the inequality, why it should hold? Can anyone help?

You are indeed correct. What you describe is a more restricted case of the proof that a "Laplace mechanism" is $$\varepsilon$$-differentially private. Just so that we're all on the same page, here's a simplified definition of $$\varepsilon$$-differentially private (adapted from Dwork and Roth, The Algorithmic Foundations of Differential Privacy, Definition 2.4):
"A randomized algorithm $$\mathcal{M}$$ with domain $$\mathbb{N}^{|\mathcal{X}|}$$ is $$\varepsilon$$-differentially private if for all $$\mathcal{S} \subseteq \text{Im}(\mathcal{M})$$ and all $$x, x' \in \mathbb{N}^{|\mathcal{X}|}$$ where $$\lVert x - x' \rVert_1 \leq 1$$, $$\frac{P(\mathcal{M}(x) \in \mathcal{S})}{P(\mathcal{M}(x') \in \mathcal{S})} \leq \exp(\varepsilon)."$$
Intuitively, for a pair of datasets $$x, x'$$ of size $$|\mathcal{X}|$$ containing categorical data that differ in at most one location, "running" $$\mathcal{X}$$ on each dataset does not leak "too much information" (i.e., bounded above as a function of $$\varepsilon$$) about the difference between the two datasets. To check your intuition, consider what $$\varepsilon$$ would be if $$x = x'$$, and/or what it means for the LHS of the inequality to be large.
So, we know that $$\mathcal{M}(x) \triangleq \langle q, x \rangle + z$$ where $$z$$ is drawn from a mean-zero Laplace distribution with parameter $$\lambda$$. Choose any $$q$$ (conforming to your problem constraints). Using the Laplace PDF, we can write the LHS of the differential privacy definition (where $$z$$ is some realized value of a Laplace RV) as $$\frac{p_x(z)}{p_{x'}(z)} = \frac{(2\lambda)^{-1}\exp(-\lambda^{-1}|z - \langle q, x \rangle|)}{(2\lambda)^{-1}\exp(-\lambda^{-1}|z - \langle q, x' \rangle|)} = \exp\left(\lambda^{-1}(|z - \langle q, x' \rangle|-|z - \langle q, x \rangle|)\right).$$
After simplification, you will be able to bound this value from above by $$\exp(2 / \lambda)$$, which yields your desired result. I'd like to leave that as an exercise (it is mostly algebra), but feel free to ask for hints.