Going by your comment you want to do this without using determinant, so without using the characteristic polynomial. There are different ways, here is one example approach.

You can view Mv = m*v as a system of equations in 3 variables x,y,m where m is the eigenvalue.

x + 2y = mx

3x + 4y = my

The eigenvalues are the values of m for which the system has a solution other than x = y = 0. Consider that system as

(1-m)x + 2y = 0

3x + (4-m)y = 0

And you can reduce it to asking the question, for what values of m does that system have a non-zero solution in x,y. That will only happen when row reduction produces a row of 0, since otherwise back substitution would yield only the single solution x = y = 0.

You can do row reduction in cases. For example to start, since row reduction depends on the value of m-1 consider the cases m = 1 or m != 1, With m != 1, you can divide symbolically by m-1 and continue to the next step, etc.