If the number of murders per year is large enough you can suppose that the distribution of your random variable $X = \text{number of murders}$ is Poissonian distributed $X\sim\text{Pois}(\lambda)$.
Having many observation, one for every year, you have a set of random variables $X_1\sim\text{Pois}(\lambda_1),\ldots, X_n\sim\text{Pois}(\lambda_n)$.
You want to test if there is a significant difference between $X_i$ and $X_j$ knowing the observation $x_i$ and $x_j$. So define the null hypothesis as $\lambda_i = \lambda_j$.
We have define the statistical model. Now you have to define your test-statistics. One natural choice is $x_i - x_j$. This is the difference between two Poissonian distribution, and it is know to be the Skellam distribution with pmf $f_{X_i-X_j}$. I guess you want to do a two-side test since you are interested in the difference in both directions. You just have to compute the p-value, since the Skellam distribution is symmetric under the null hypothesis:
$$\text{p-value} = 2\sum_{x\geq |x_i-x_j|} f_{X_i-X_j}(x, \lambda, \lambda)$$
One thing missing here is the value of $\lambda=\lambda_i=\lambda_j$, I guess you can just take the mean of $x_i$ and $x_j$.
Alternatively you can use a more general approach and use the profiled likelihood ratio.
$$\Lambda = -2 \log\frac{\sup_\lambda\text{Pois}(x_i|\lambda)\text{Pois}(x_j|\lambda)}{\sup_{\lambda,\Delta_1,\Delta_2}\text{Pois}(x_i|\lambda+\Delta_i)\text{Pois}(x_2|\lambda+\Delta_j)}$$
and then compute the p-value, using the fact that, asymptotically, $\Lambda$ is distributed as a $\chi^2$-distribution with 1 degree of freedom. After some math you can simplify the expression:
$$\Lambda = -2(x_i + x_j)\log\left(\frac{x_i + x_j}{2}\right) + 2x_i \log x_i + 2 x_j \log x_j$$
Here the histogram for $\Lambda$ generating 10k pseudo-experiment under the null-hypothesis
With these tests I am ignoring the information from the correlation between the years, but to take into account them you need a model describing the behaviour of the murders across the years.
