Correlated Poisson Distribution

You haven't given enough information to say what their joint distribution is. But maybe the simplest possibility is this: \begin{align} X_1 & = Y_1 + Y_2 \\ X_2 & = Y_2 + Y_3 \end{align} where $Y_1,Y_2,Y_3$ are independent Poisson-distributed random variables with expectations $\mu_1,\mu_2,\mu_3$, and of course $\mu_1+\mu_2=\lambda_1$ and $\mu_2+\mu_3=\lambda_2$. Recall that the variance of a Poisson-distributed random variable is the same as its expected value. Then the correlation is $\rho$ if $$ \operatorname{cov}(X_1,X_1) = \rho\sqrt{\operatorname{var}(X_1)\operatorname{var}(X_2)}=\rho\sqrt{\lambda_1\lambda_2}. $$ The covariance is $$ \operatorname{cov}(X_1,X_1) = \operatorname{cov}(Y_1+Y_2,Y_2+Y_3) = \operatorname{var}(Y_2)=\mu_2. $$ From there you can think about possible values of $\mu_1,\mu_2,\mu_3$ as functions of $\lambda_1,\lambda_2$.

Consider this model that could generate correlated Poisson variables. Let $Y$, $Y_1$ and $Y_2$ be three independent Poisson variable with parameters $r$, $\lambda_1$ and $\lambda_2$. Let $$X_i=Y_i+Y$$ for $i=1,2$. Then $X_1$ and $X_2$ are both Poisson with parameters $\lambda_1$ and $\lambda_2$. They have the correlation $$\rho=\frac{r}{\sqrt{(\lambda_1+r)(\lambda_2+r)}}$$ Now the joint distribution can be derived as $$P[X_1=i,X_2=j]=e^{-(r+\lambda_1+\lambda_2)}\sum_{k=0}^{i\wedge j}\frac{r^k}{k!}\frac{\lambda_1^{(i-k)}}{(i-k)!}\frac{\lambda_2^{(j-k)}}{(j-k)!}$$ The case for a bivariate Poisson process is immediate from here.

You could look at the Johnson and Kotz book on multivariate discrete distributions for more information (this construction of a bivariate Poisson distribution is not unique). Also, it has the drawback that $\rho \in [0, \min(\lambda_1, \lambda_2)/\sqrt{\lambda_1\lambda_2} ]$ when $\lambda_1 \neq \lambda_2$ as discussed by Genest et al. 2018.