An answer based on J.D.'s hint (and essentially equivalent to Sebastian's excellent solution):
Let $r(t)=t^4+t+1$. Then $r(t)^{4^i}=t^{4^{i+1}}+t^{4^i}+1$ for all $i$. Therefore $ \begin{aligned} r(t)+&r(t)^4+r(t)^{16}+r(t)^{64}+r(t)^{256}=\\&=5\cdot1+(t+t^4+t^{16}+t^{64}+t^{256})+(t+t^4+t^{16}+t^{64}+t^{256})^4\\&=1+t+t^{1024}.\end{aligned} $ But clearly the l.h.s. is divisible by $r(t)$, so is the r.h.s., and hence it cannot be irreducible. Not really brute force, but not entirely obvious, unless you have spent some time with finite fields.
Yet another way, but not the recommended way :-), of proving the claim is the following. We know that $q(t)=t^{1024}+t$ has the property that $q(u+v)=q(u)+q(v)$ for all $u,v$ in an algebraic closure of $F_2$. But we also know that $q(u)=0$ if and only if $u\in F_{1024}$. Now if we assume contrariwise that $P(t)$ is irreducible, and $y$ is one of its roots, then $y$ would generate the field $F_{2^{1024}}$. This field is Galois over $F_2$, so it must contain all the roots of $P(t)$ (being a normal extension). But the earlier observations imply that all the elements of the form $y+u$, with $u\in F_{1024}=F_{2^{10}}$ must also be zeros of $P(t)=q(t)+1$, because $P(y+u)=q(y+u)+1=q(y)+q(u)+1=q(y)+1=P(y)=0.$ Thus also $u=(y+u)+y$ belongs to the splitting field $F_{2^{1024}}$ of $P(t)$. Here $u\in F_{2^{10}}$ was arbitrary, so $F_{2^{10}}\subseteq F_{2^{1024}}$. But this is a contradiction, because $10\nmid 1024$.
This kind of manipulations are not uncommon when playing with linearized (and affine) polynomials in $F_p[t]$, i.e. polynomials with terms of degrees that are powers of $p$ only.