Let $k$ be a Natural number. I want to show that for $N \gg k$ there does not exist a subalgebra $\alpha \subset \mathrm{End}( \mathbb{C}^N)$ of dimension $\ge N$ such that every $\varphi \in \alpha$ has minimal polynomial of degree $\le k$ and no elements of rank $1$ (hence also no elements of corank $1$).
My idea is that such an algebra due to the hypothesis on the minimal polynomial must contain two non commuting endomorphisms. My intuition is that the minimal polynomial of the product should be more complicated (maybe one can prove that its degree increases?) . But I couldn`t work it out.
Notice that the bound on the dimension of the algebra may be far from being optimal and maybe even indipendent from $N$ a priori. My understanding of the problem is so limited that to me it can even be $k^2$ or being completely false. As far as I understand the hypothesys on the rank may be useless.
Ok this took me a while but your claim is actually false. Let $N = 4t$, and first consider the rng(ring without identity) $$B = \begin{bmatrix}0 & X & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & X \\ 0 & 0 & 0 & 0 \end{bmatrix}, X \in M_t(\mathbb{C}).$$ Note that the shape of $B$ guarantees that the product of any two elements of $B$ is $0$. Furthermore, any nonzero element of $B$ has rank at least $2$.
Now let $A = \mathbb{C} I + B$. Then $A$ is an algebra, every element of $A$ has minimal polynomial of degree at most $2$, and no element of $A$ has rank $1$. But the dimension of $A$ is $t^2 + 1 > N$, for $t\ge 4$.