Suppose $X~Poisson(\mu_1)$ and $Y~Poisson(\mu_2)$ and $X$ and $Y$ are independent.
Then $p(x,y;\mu_1, \mu_2)=(exp(-\mu_1-\mu_2)\cdot \mu_1^x \mu_2^y/(x!y!)$.
How do I obtain a general expression for the probability of $m$ where $x+y=m$ (where $m$ is a non-negative integer).
On
You have $P(X=x,Y=y),$ the joint distribution of $X$ and $Y.$ This gives the probability that $(X,Y)$ is any ordered pair. Now you need to get the probability that $X+Y = z$ for any $z.$ Start by computing this for some value of $z.$ How about $z=0.$ There's only one way for $X+Y=0,$ namely $(X,Y) = (0,0),$ so $$P(Z=0) = P(X=0,Y=0) = e^{-(\mu_1+\mu_2)}.$$ That one's too easy. We'll skip to $z=2$. For this we can either have $(X,Y) = (2,0)$ or $(1,1)$ or $(0,2)$ so $$ P(Z=2) = P(X=2,Y=0) + P(X=1,Y=1)+ P(X=0,Y=2) \\=e^{-(\mu_1+\mu_2)}(\frac{1}{2}\mu_1^2 + \mu_1\mu_2 + \frac{1}{2}\mu_2^2) =e^{-(\mu_1+\mu_2)}\frac{(\mu_1+\mu_2)^2}{2}.$$
Now do it for three. If you think about how the answer looks you should be able to guess the right answer.
In any event, generalizing the above example to an arbitrary $z,$ we can write $$ P(Z=z) = \sum_{n=0}^zP(X_1=z-n,X_2=n)\\= \sum_{n=0}^\infty e^{-(\mu_1+\mu_2)}\frac{\mu_1^{z-n}\mu_2^n}{(z-n)!n!}$$ so you just need to compute this sum. (Again you should be able to get some ideas from doing $z=2$ and $z=3$... also remembering things about the binomial formula will help.)
Define $M = X + Y$. I like to use upper case letters for random variables. Then:
$P(M=m)=P(X+Y=m)=\sum_xP(Y=m-x|X=x) P(X=x)$.
I hope the above helps give you a start with the problem.