Rewrite the terms with a common base.

The equation is $49^{x+1} + 7^{2x} = 350$.
Since $49 = 7^2$, we can rewrite $49^{x+1}$ as $(7^2)^{x+1}$.

$$(7^2)^{x+1} + 7^{2x} = 350$$

$$7^{2(x+1)} + 7^{2x} = 350$$ $$7^{2x+2} + 7^{2x} = 350$$

$$7^{2x} \cdot 7^2 + 7^{2x} = 350$$ $$7^{2x} \cdot 49 + 7^{2x} = 350$$

$$7^{2x}(49 + 1) = 350$$ $$7^{2x}(50) = 350$$

$$7^{2x} = \frac{350}{50}$$ $$7^{2x} = 7$$

Since $7 = 7^1$, we can write:

$$7^{2x} = 7^1$$

For the equality to hold, the exponents must be equal.

$$2x = 1$$

$$x = \frac{1}{2}$$

The value of $x$ is $\boxed{\frac{1}{2}}$.