How can I prove that $f'(x)=1+\left[f(x)\right]^2$ has no solution over $\mathbb{R}$ without solving the equation explicitly?

One can often estimate the existence interval by solving a simpler differential equation and obtaining a lower bound for the solution.

In your case: Assume that $f$ solves the differential equation on the interval $[a, b]$ with $f(a) > 0$ (the case $f(a) < 0$ can be handled similarly). Then $f$ is strictly positive on the interval, and $$ f'(x) = 1 + f(x)^2 > f(x)^2 $$ which implies that $$ b - a < \int_a^b \frac{f'(x)}{f(x)^2} \, dx = \frac{1}{f(a)} - \frac{1}{f(b)} < \frac{1}{f(a)} $$ and shows that $b$ cannot be arbitrarily large.

More concretely: Let $f$ be the solution with $f(0) = 0$. Then $f'(x) \ge 1$ for $x \ge 0$ so that $f(1) \ge 1$. Applying the above with $a=1$ shows that $$ b < 1 + \frac{1}{f(1)} \le 2 \, , $$ i.e. no solution exists on the interval $[0, 2]$.

Suppose $y(x)$ is an everywhere differentiable function which satisfies the differential equation $y'=1+y^2$.

Our goal is to derive a contradiction (without explicitly solving the ODE).

From $y'=1+y^2$, it follows that $y'\ge 1$ for all $x\in\mathbb{R}$, hence $y$ is increasing.

If $y$ is bounded above, then since $y$ is increasing, we would have $$ \lim_{x\to\infty}y'(x)=0 $$ contrary to $y'\ge 1$ for all $x\in\mathbb{R}$.

Similarly, if $y$ is bounded below, then since $y$ is increasing, we would have $$ \lim_{x\to -\infty}y'(x)=0 $$ contrary to $y'\ge 1$ for all $x\in\mathbb{R}$.

Hence the range of $y$ is equal to $\mathbb{R}$.

Let $a\in\mathbb{R}$ be such that $y(a)=1$ and let $b > a$. \begin{align*} \text{Then}\;\;& y'=1+y^2 \\[4pt] \implies\;& \frac{1}{1+y^2}\,dy=dx \\[4pt] \implies\;& \int_{y(a)}^{y(b)}\frac{1}{1+y^2}\,dy=\int_a^b 1\,dx \\[4pt] \implies\;& \int_1^{y(b)}\frac{1}{1+y^2}\,dy=b-a \\[4pt] \implies\;& \lim_{b\to\infty}\left(\int_1^{y(b)}\frac{1}{1+y^2}\,dy\right)=\infty \\[4pt] \implies\;& \int_1^\infty \frac{1}{1+y^2}\,dy=\infty \\[4pt] \end{align*} contradiction, since $$ \int_1^\infty \frac{1}{1+y^2}\,dy < \int_1^\infty \frac{1}{y^2}\,dy = 1 $$