If you already have the moment generating function, differentiating a few times will show you the pattern given in wikipedia: Touchard polynomials. They satisfy $E[X^n] = \lambda(1 + \tfrac{d}{d\lambda})E[X^{n-1}].$
A simple way to calculate $E[X^4] = e^{-\lambda}\sum_{k=0}^\infty \frac{k^4\lambda^k}{k!}$ is by differentiating:
$$ d^4/d\lambda^4 \lambda^k = e^\lambda = \sum_{k=0}^\infty \frac{(k^4 - 6k^3 - 11k^2 + 6k)\lambda^{k-4}}{k!}\\ = \frac{1}{\lambda^4}(E[X^4] - 6E[X^3] - 11E[X^2] + 6E[X]) $$$$ \frac{d^4}{d\lambda^4} e^{\lambda} = e^\lambda = \sum_{k=0}^\infty \frac{(k^4 - 6k^3 + 11k^2 - 6k)\lambda^{k-4}}{k!}\\ = \frac{e^{\lambda}}{\lambda^4}(E[X^4] - 6E[X^3] + 11E[X^2] - 6E[X]) $$ This gives $E[X^4] = \lambda^4 + 6E[X^3] - 11E[X^2] + 6E[X]$ and, in the same way, $E[X^3] = \lambda^3 + 3E[X^2] - 2E[X] = \lambda^3 + 3\lambda^2 + \lambda$ since $E[X^2] = \lambda + \lambda^2$. Thus
$$ E[X^4] = \lambda^4 + 6\lambda^3 + 18\lambda^2 + 6\lambda - 11\lambda^2 - 5\lambda = \lambda^4 + 6\lambda^3 + 7\lambda^2 + \lambda. $$