Can $\pi$ be a root of a polynomial under special cases? What if we consider polynomials whose coefficients are either rational or $e$, that is, a polynomial in $\mathbb{Q} \cup \{e\}$ with $\pi$ as a root. 
Can this happen? Does it matter if we change rational to integer in the definition?
 A: The answer is almost certainly no, but no proof is known.  In fact, it is not even known whether $\pi  e$ is rational.    Schanuel's conjecture implies that $\pi$ and $e$ are algebraically independent, and therefore that the answer to your question is no.  
A: A real/complex number is a root of a polynomial with rational coefficients if and only if it is a root of a polynomial with integer coefficients: just multiply by a suitable integer to clear denominators to get an integer polynomial.
(However, it is not true that a complex number is a root of a monic polynomial with integer coefficients if and only if it is a root of a monic polynomial with rational coefficients; the former are called algebraic integers, and form a proper subset of the algebraic numbers. For example, $\frac{1}{2}$ is certainly the root of a monic polynomial with rational coefficients, e.g., $x - \frac{1}{2}$, but it's not the root of any monic polynomial with integer coefficients.)
Now, of course $\pi$ is a root of many polynomials with real coefficients: it's a root of $x-\pi$, after all, and so on. 
Whether $\pi$ is a root of a polynomial with coefficients in $\mathbb{Q}[e]$ or in $\mathbb{Q}(e)$, however, is unknown. The notion you want is that of algebraic independence. A set of real numbers $\{r_1,\ldots,r_n\}$ is said to be "algebraically independent" if there does not exist a nonzero polynomial $p$ in $\mathbb{Q}[x_1,\ldots,x_n]$ such $p(r_1,\ldots,r_n)=0$. It is unknown whether $\{\pi,e\}$ is algebraically independent. (It's not even known whether $\pi+e$ is irrational!)
However, it is known that $\pi$ and $e^{\pi}$ are algebraically independent.
