Functional equation with the property $P(x+1)=P(x)+2x+1$ 
Find all polynomials with real-valued coefficients for which the following property $$P(x+1)=P(x)+2x+1$$ holds for all $x \in \mathbb{R}.$

This seems to be a functional equation, so the initial approach would be to try some values. For $x=0$ we would have that $$P(1)=P(0)+1.$$
For $x=1$ $$P(2)=P(1)+3.$$
For $x=2$ $$P(3) = P(2)+5.$$
Now if I would know something about $P(0)$ that would seem to be helpful. If I assume that $P(0)=0$ I would get that
$$P(1)=1$$ $$P(2)=4$$ $$P(3)=9.$$
From here it would seem that $P(n)=n^2,$  however I’m not sure if I can make the assumption that $P(0)=0$ or can I?

Suppose $\deg P=n$. Then we can write
$$P(x)=\sum_{k=0}^{n}a_kx^k.$$
From the functional equation,
$$\sum_{k=0}^{n}a_k(x+1)^k=1+2x+\sum_{k=0}^{n}a_kx^k.$$
Using the property of binomial coefficients $$m>n\Rightarrow {n\choose m}=0,\tag{1}$$
we have $$P(x+1)=\sum_{k=0}^{n}a_k\sum_{j=0}^{n}{k\choose j}x^j=\sum_{j=0}^{n}\alpha_jx^j,$$
where $$\alpha_j=\sum_{k=0}^{n}{k\choose j}a_k.\tag{2}$$
Comparing coefficients, we have that 
$$\begin{align}
\alpha_0&=1+a_0,\\
\alpha_1&=2+a_1,\\
\alpha_j&=a_j\qquad j\ge 2.
\end{align}$$
Then from $(2)$, we have 
$${0\choose 2}a_0+{1\choose 2}a_1+{2\choose 2}a_2+{3\choose 2}a_3+...+{n\choose 2}a_n=a_2.\tag{3}$$
We have that $(3)$ implies $a_j=0$ for all $j\ge 3$, because ${0\choose 2}={1\choose 2}=0$ from $(1)$. Therefore, $\deg P=2$ and we have 
$$\alpha_j=\sum_{k=0}^{2}{k\choose j}a_k.$$
Thus 
$$\begin{align}
1+a_0=&\alpha_0=a_0+a_1+a_2\\
2+a_1=&\alpha_1=a_1+2a_2.
\end{align}$$
Thus we are left with the system of equations
$$\begin{align}
1=&a_1+a_2\\
2=&2a_2,
\end{align}$$
which has the solution $(a_0,a_1,a_2)=(a_0,0,1)$, meaning that $a_0$ can be any constant. Thus we have
$$P(x)=x^2+a_0$$
for any $a_0\in\Bbb R$.
