I tried converting to exponential, $e^{1+x^2} -x -1 = 0$ then finding a contradiction but didn't find one. Then I differentiated to find a minima $> 0$.
i.e solution of $2xe^{1+x^2} -1 = 0$ but didn't find a solution as the term $2xe^{1+x^2}$ doesn't equal to $1$ at any point $x$. Again even if there was any it still would have been a local minima.
The derivatives of the two functions are $\frac1{1+x}$ and $2x$. The latter derivative is greater than the former beyond $x_0=0.366\dots$, and we also have $1+x_0^2>\ln(1+x_0)$. This shows that there are no intersections in $[x_0,\infty)$.
In the interval $(-1,x_0]$, $1+x^2\ge1$ but $\ln(1+x)<1$. There are no intersections in this interval either. So the two graphs never intersect.
Let $f(x)=1+x^2-\ln(1+x).$
Thus, $$f'(x)=2x-\frac{1}{1+x},$$ which gives $$x_{min}=\frac{\sqrt3-1}{2}$$ and easy to check that: $$f\left(\frac{\sqrt3-1}{2}\right)>0.%$$
$e^t \ge 1+t$ holds for all real numbers $t$, see for example Simplest or nicest proof that $1+x \le e^x$.
It follows that $$ e^{1+x^2} -x-1 \ge 2+x^2-x-1 = \left(x - \frac 12 \right)^2 + \frac 34 > 0 $$ so that $e^{1+x^2} > 1+x$ for all $x \in \Bbb R$.
An option:
$x>-1;$
Assume the contrary.
$e^{1+x^2}=x+1; $
1)$-1<x \le 0;$
No solution since
$0< RHS \le 1,$ and $LHS >e;$
2)$0<x \le 1;$
LHS$\ge e$, RHS $\le 2;$
3)$x>1;$
LHS: $1+x^2 + T(x^2)$ (expanding),
where $T >0;$
RHS: $1+x;$
LHS > RHS for $x >1.$
