I am trying to solve the following equation to get the eigenvalues of this (n + 1 $\times$ n + 1) matrix m:
$$m = \begin{bmatrix} a & a1^T \\ a1 & bJ \end{bmatrix}$$ Where: $$ a, b \in R, 1^T = [1.....1]^T $$ and J is a (n $\times$ n) matrix of all $1$s.
so far, I have done the following(I is the identity matrix here):
$$det(m) = \begin{bmatrix} a - \lambda & a1^T \\ a1 & bJ - \lambda I \end{bmatrix} = 0 \\ abJ - a\lambda I - \lambda bJ + \lambda^2 I - a1^Ta1 = 0 \\ \lambda^2I - \lambda(aI + bJ) + abJ - na^2 = 0 $$ because $$a1^Ta1 = [a....a][a....a]^T = na^2$$
How do I proceed from here?
Your computation of the determinant is not correct: you are treating $J$ as if it were a scalar, when it is a matrix. It should be obvious since you get a degree two equation on $\lambda$ instead of degree $n$.
For this matrix, it is easier to look at the eigenvalues directly. I will assume $a\ne b$ and both nonzero, since it is the non-trivial case.
The equation $mx=\lambda x$ gives the equations $$\tag1 a\sum_{j=1}^n x_j=\lambda x_1,\ \ ax_1+b\sum_{j=2}^n x_j=\lambda x_2=\cdots=\lambda x_n. $$ Consider first the case $\lambda=0$. Now the equations are $$ \sum_{j=1}^n x_j=0,\ \ ax_1+b\sum_{j=2}^n x_j=0. $$ Solving for $x_1$ in the first equation we get $(b-a)\sum_{j=2}^n x_j=0$. Then $x_1=-\sum_{j=2}^n x_j=0$. So the eigenspace for $\lambda=0$ is the subspace $$ \{x:\ x_1=0,\ \sum_{j=2}^n x_j=0\}, $$ which has dimension $n-2$.
Now consider the case $\lambda\ne0$. We obtain from $(1)$ that $x_2=\ldots=x_n$. They cannot be zero, because $(1)$ would become $ax_1=\lambda x_1$ and $ax_1=0$, a contradiction. So assume that $x_2=1$. Now $(1)$ becomes $$ (n-1)=(\lambda-a)x_1, \ \ \ ax_1+b(n-1)=\lambda; $$ solving for $\lambda$ we obtain a quadratic equation on $\lambda$ that will give us the other two eigenvalues. Solving the quadratic quickly gives me $$ \lambda =\frac{a+b(n-1)\pm\sqrt{[b(n-1)+a]^2-4(n-1)(b-1)a}}{2} $$